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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The slope of line which belongs to family (1+ l) x + (1-l)y + 2(1-l) = 0 and makes shortest intercept on `x^2 = 4y - 4` |
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Answer» `(1+l)x+(l-1)y+2(1-l)=0` `x+xl+ly-y+2+2l=0` `(x-y+2)+l(x+y-2)=0` `x-y+2=0-(1)` `x+y-2=0-(2)` adding equation 1 and 2 x=0,y=0 `x^2=4(y-1)` `F(0,2)` `1/(FA)+1/(FB)=2/(FC)` option c is correct |
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| 2. |
. The shortest distance from the point (2, -7) to circle `x^2+y^2-14x-10y-151=0` |
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Answer» False Given circle is `x^(2)+y^(2)-14x-10y-151=0` `therefore` Centre=(7,5) and Radius=`sqrt(49+25+151)=sqrt(225)=15` So, the distances between th point (2,-7) and centre of the circle is given by `d_(1)=sqrt((2-7)^(2)+(-7-5)^(2))` `=sqrt(25+144)=sqrt(169)=13` `therefore` Shortest distance,d=`abs(13-15)=2` |
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| 3. |
A tangent to a hyperbola `x^2/a^2 - y^2/b^2 = 1` intercepts a length of unity from each of the coordinate axes, then the point `(a, b)` lies on rectangular hyperbola |
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Answer» let the eqn of hyperbola be `H : x^2/a^2 -y^2/b^2 = 1` let the point on it be `(a sec theta, b tan theta)` the equation of tangent will be `bxsec theta - ay tan theta = ab` putting y=0 `bxsec theta = ab ` `x= a cos theta = 1` putting x=0 `-aytan thea = ab` so, `y = bcos theta=1` as `acos theta = 1` `a = sec theta` as `-bcos theta = 1` `b = -tan theta` as we know the identity `sec^2 theta - tan^2 theta = 1` `a^2 - (-b)^2 =1` `a^2-b^2=1 ` `x^2 - y^2=1 ` |
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| 4. |
If the focus of a parabola is (0,-3) and its directrix is y=3, then its equation isA. `x^(2)=-12y`B. `x^(2)=12y`C. `y^(2)=-12x`D. `y^(2)=12x` |
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Answer» Correct Answer - A Given that, focus of parabola at F (0,-3) and equation of directrix is y=3 Let any point on the parabola is P (x,y). Then, PF=`abs(y-3)` `rArr sqrt((x-0)^(2)+(y-3)^(2))=abs(y-3)` `rArrx^(2)+y^(2)+6y+9=y^(2)-6y+9` `rArrx^(2)+12y=0` `rArrx^(2)=-12y` |
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| 5. |
Find the equation of the circle which circumscribes the triangle formedby the line: ` y=x+2,3y=4x a n d 2y=3x` |
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Answer» Given equations of line are y=x+2…(i) 3y=4x….(ii) 2y=3x ….(iii) From Eqs. (i) and (ii) `(4x)/3=x+2` `rArr4x=3x+6` `rArr x=6` On putting x=6 in Eq.(i) we get y=8 `therefore` Point,A=(6,8) From Eqs. (i) and (iii), `(3x)/2=x+2` `rArr3x=2x+4rArrx=4` When x=4 then y=6 `therefore` Point, B=(4,6) From Eqs. (ii) and (iii) `x_(1)=0_(1),y=0` Now, C=(0,0) Let the equation of circle is `x^(2)+y^(2)+2gxn+2fy+c=0` since,the points A(6,8),B(4,6) and C(0,0) lie on this circle 36+64+12g+16f+c=0 `rArr 12g+16f+c=-100....(iv)` and 16+36+8g+12f+c=0...(v) `rArr 8g+12f+c=-52`....(vi) `rArr c=0` From Eqs. (iv) , (v) and (vi), 12g+16f=-100 `rArr3g+4f+25=0` `rArr2g+3f+13=0` `rArrg/(+52-75)=f/(50-39)=1/(9-8)` `rArrg/(-23)=f/11=1/1` `rArrg=-23,f=11` So, the equation of the circle is `x^(2)+y^(2)-46x+22y+0=0` `rArrx^(2)+y^(2)-46x+22y=0` |
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| 6. |
Find the equation of the parabola whose focus is at (-1,-2) and thedirectrix the line `x-2y+3=0` |
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Answer» Given that, focus at F(-1,-2) and dirctrix is x-2y+3=0 Let any point on the parabola be (x,y) `PF=abs((x-2y+3)/(sqrt(1+4)))` `rArr (x+1)^(2)+(y+2)^(2)=((x-2y+3)^(2))/5` `rArr5[x^(2)+2x+1+y^(2)+4y+4]=x^(2)+4y^(2)+9-4xy-12y+6x` `rArr4x^(2)+y^(2)+4x+32y+16=0` |
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| 7. |
Find the equation of the circle having `(1,-2)`as its centre and passing through the intersectionof the lines `3x+y=14a d n2x+5y=18.` |
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Answer» Given that, centre of the circle is (1,-2) and the circle passing through the lines 3x+y= 14….(i) and 2x+5y=18…(ii) From Eq. (i) y=14-3x pu in Eq. (ii), we get 2x+70-15x=18 `rArr` -13x=-52`rArr`x=4 Now, x=4 put in Eq. (i) we get 12+y=14`rArr`=y=2 Since, point (4, 2) lie on these lines also lies on the circle. Radius of the circle =`sqrt((4-1)^(2)+(2+2)^(2))` `=sqrt(9+16)=5` Now equation of the circle is `(x-1)^(2)+(y+2)^(2)=5^(2)` `rArr x^(2)-2x+1+y^(2)+4y+4=25` `rArr x^(2)+y^(2)-2x+4y-20=0` |
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| 8. |
Find the equations of the hyperbola satisfying the given conditions :Foci `(0,+-sqrt(10))`, passing through (2,3) |
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Answer» Here, Foci of hyperbola `= (+-sqrt10,0)` That means the transverse axis of the hyperbola is `Y`-axis. So, the equation will be of the type, `y^2/a^2-x^2/b^2 = 1->(1)` Also, `c = sqrt10` In a hyperbola, `c^2 = a^2+b^2` Putting value of `c`, `=> (sqrt10)^2 = a^2+ b^2` `=>b^2 = 10-a^2` Putting vallue of `b^2` in (1), `y^2/a^2-x^2/10-a^2 = 1->(2)` As hyperbola is passing through `(2,3)`, `x=2` and `y=3` should satisfy (2). `So, 9/a^2-4/(10-a^2) = 1` `=>9(10-a^2)-4a^2 = a^2(10-a^2)` `=>90-9a^2-4a^2 = 10a^2 - a^4` `=>a^4-23a^2+90 = 0` `=>a^4-18a^2-5a^2+90 = 0` `=>(a^2-18)(a^2-5) = 0` `=> a^2 = 18 and a^2 = 5` But, `a^2` can not be greater than `c^2`. `:.a^2 = 5` `b^2 = 10 - 5 = 5` Putting values of `a^2` and `b^2` in (1), `y^2/5-x^2/5 = 1` |
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| 9. |
Find the equation of the circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line `y = x-1` |
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Answer» Let equation of circle be `(x-h)^(2)+(y-k)^(2)=r^(2)…..(i)` `rArr (x-h)^(2)+(y-k)^(2)=9` Given that, centre (h,k) lies on the line y=x-1 i.e., k=h-1 Now, the circle passes through the point (7,3). `(7-h)^(2)+(3-k)^(3)=9` `rArr 49-14h+h^(2)+9-6k+k^(2)=9` `rArr h^(2)+k^(2)-14h-6k+49=0`.....(iii) On putting k=h-1 in Eq. (iii) we get `h^(2)+(h-1)^(2)-14h-6(h-1)+49=0` `rArr 2h^(2)-22h+56=0` `rArrh^(2)-11h+28=0` `rArr h^(2)-7h-4h+28=0` `rArrh(h-7)-4(h-7)=0` `rArr(h-7)(h-4)=0` `therefore h=4,7` When h=7, then k=7-1=6 `therefore` Centre(7,6) When h=4, then k=3 `therefore` Centre-(4,3) So, the equation of circle when centre (7,6) is `(x-7)^(2)+(y-6)^(2)=9` `rArr x^(2)-14x+49+y^(2)-12y+36=9` `rArr x^(2)+y^(2)-14x-12y+76=0` When centre (4,3), then the equation of the circle is `(x-4)^(2)+(y-3)^(2)=9` `therefore x^(2)-8x+16+y^(2)-6y+9=9` `rArr x^(2)+^(2)-8x-6y+16=0` |
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| 10. |
The equation of the hyperbola with vertices at (0,`pm`6) and eccentricity 5/3 is….. And its foci are ……. |
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Answer» Let the equation of the hyperbola be -`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` Then vertices=(0,`pm`b)=(0,`pm`6) `therefore b=6 and e=5//3` `becausee=sqrt(1+(a^(2))/(b^(2)))rArr25/9=1+(a^(2))/36` `rArr (25-9)/9=(a^(2))/36rArr16=(a^(2))/4rArra^(2)=48` So, the equation of hyperbola is `(-x^(2))/48+(y^(2))/36=1rArr(y^(2))/36-(x^(2))/48=1` `because` Foci=(0,`pm`be)=(0,`pm5/3xx6`)=(0,`pm`10) |
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| 11. |
If the parabola `y^2=4a x `passes through the point (3,2) then find the length of its latusrectum.A. `2/3`B. `4/3`C. `1/3`D. 4 |
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Answer» Correct Answer - B Given that, parabola is `y^(2)=4ax` `therefore` Length of latusrectum=4a Since, the parabola passes through the point (3,2) Then, 4=4a(3) `rArr a=1//3` `therefore` 4a=4/3 |
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| 12. |
If the line `l x+m y+n=0`touches the parabola `y^2=4a x ,`prove that `ln=a m^2` |
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Answer» True Given equation of a line is lx+my+n=0 and parabola`y^(2)`=4ax From Eq. (i), x=`-((my+n)/l)`put in Eq. (ii) we get `y^(2)=-(4a(my+n))/l` `rArr ly^(2)=-4amy-4ax` `rArrly^(2)+3amy+4an=0` For tangent, D=0 `rArr16a^(2)m^(2)=4lxx4an` `rArr 16a^(2)m^(2)=16anl` `rArr am^(2)=nl` |
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| 13. |
A circle of radius R touches externally a set of 12 circles each of radius `r` surrounding it. Each one of the smaller circles touches two circles of the set. Then `R/r=sqrtm+sqrtn-1,` where `m,n in N and m+n` is |
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Answer» Sum of interior angles of 12 sides polygon is=`(n-2)pi` `=10pi` `(10pi)/12=2(pi/2-theta/2)` `(5pi)/6=pi-theta` `theta=pi/6=30^0` `(2r)^2=R^2+R^2-2Rcostheta` `4r^2=2R(1-cos30^0)` `R^2/r^2=1/(1-sqrt3/2)` `=4/(2-sqrt3)` `=4(2+sqrt3)` `R/r=sqrt2*sqrt(4+2sqrt3` `=sqrt2*(sqrt3+1)` `=sqrt6+sqrt2` m=6,n=2 m+n=6. |
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| 14. |
The difference between the radii of the largest and smallest circles which have their centres on the circumference of the circle `x^2 + y^2 + 2x + 4y -4 = 0` and passes through point (a,b) lying outside the circle is : |
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Answer» `x^2+y^2+2x+4y-4=0` `(x+1)^2+(y+2)^2-1-4-4=0` `(x+1)^2+(y+2)^2=3^2` AB is a diameter `r_m=AB+r_s` `r_m-r_s=AB=2*3=6`. |
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| 15. |
If e is eccentricity of the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`(where,a`lt`b), thenA. `b^(2)=a^(2)(1-e^(2))`B. `a^(2)=b^(2)(1-e^(2))`C. `a^(2)=b^(2)(e^(2)-1)`D. `b^(2)=a^(2)(e^(2)-1)` |
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Answer» Correct Answer - B Given that, `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 altb` We know that, `e=sqrt(1-(a^(2))/(b^(2)))rArre^(2)=((b^(2)-a^(2)))/(b^(2))` `rArr b^(2)e^(2)=b^(2)=a^(2)` `rArra^(2)=b^(2)(1-e^(2))` |
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| 16. |
If lx + my -1 = 0 touches the circle `x^2 + y^2 = a^2` then the point (l,m) lies on the circle |
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Answer» C(0,0),r=a `d=|(0+0-1)/sqrt(l^2+m^2)|=a` `1/(l^2+m^2)=a^2` `l^2+m^2=1/a^2` `l^2+m^2=a^(-2)` `x^2+y^2=a^(-2)` option C is correct. |
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| 17. |
The locus of the centre of a circle touching the circle `x^2 + y^2 - 4y -2x = 2sqrt3 - 1` internally and tangents on which from (1,2) is making a `60^@` angle with each other is a circle. then integral part of its radius is |
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Answer» `x^2+y^2-4y-2x=2sqrt3-1` `(x-1)^2+(y-2)^2=1+4+2sqrt3-1` `(x-1)^2+(y-2)^2=(sqrt3+1)^2` `C_1C_2=r_1+r_2` `(h-1)^2+(k-2)^2=(sqrt3+1+r)^2` `sin30^0=r/(sqrt3+1+r)` `sqrt3+1+r=2r` `r=sqrt3+1=2.732` 2 Answer |
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| 18. |
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.`16 x^2+y^2=16` |
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Answer» Given equation is, `16x^2+y^2 = 16` We will convert this equation into standard form. `=>16/16x^2+1/16y^2 = 1` `=> x^2/1+y^2/16 =1 ` So, this is our standard equation of ellipse with, `a = 1, b = 4` `c = sqrt(b^2-a^2) = sqrt(16-1) = sqrt15` Here, as `b gt a`, major-axis will be `Y-`axis. Now, foci will be `(0,+-c) = (0,+-sqrt15).` Vertices will be `(0,+-b) = (0,+-4).` Length of major-axis ` = 2b = 8` Length of minor axis ` = 2a = 2` Eccentricity `= c/b = sqrt15/4.` Length of latus rectum ` = 2a^2/b = 2**1/4 = 1/2.` |
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| 19. |
If latus rectum of ellipse `x^2tan^2phi+y^2sec^2phi=1` is 1/2, then the value of `phi` where `phi in (0,pi)` is |
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Answer» `x^2/cos^2phi+y^2/cos^2phi=1` `a^2=cot^2phi` `b^2=cos^2phi` `1/2=(2b^2)/a=(2*cos^2phi)/cotphi` `1/2=(2cos^2phi*sinphi)/cosphi` `1/2=sin^2phi` `2phi=pi/6` `phi=pi/12` `sin(pi-2phi)=1/2` `pi-2phi=pi/6` `pi-pi/6=2phi` `2phi=5/6pi` `phi=5/12pi` option d is correct. |
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| 20. |
If the latus rectum of an ellipse is equal to the half of minor axis, then find its eccentricity. |
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Answer» Consider the equation of the ellipse is `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` `therefore`Length of major axis=2a Length of minor axis=2b and length of latusrectum=`(2b^(2))/a` Given that, `(2b^(2))/a=(2b)/2` `rArr` a=2b`rArr`b=a/2 We know that, `b^(2)=a^(2)(1-e^(2))` `rArr (a/2)^(2)=a^(2)(1-e^(2))` `rArr(a^(2))/4=a^(2)(1-e^(2))` `rArr1-e^(2)=1/4` `rArre^(2)=1-1/4` `therefore e=sqrt(3/4)=sqrt(3/2)` |
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| 21. |
Tangents are drawn from a point `P(6,sqrt(5))` to the ellipse `x^2/25+y^2/16=1`touching the ellipse in the points Q and R. The angle between PQ and PR is |
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Answer» Equation of tangent to an ellipse=>`y=mx+-sqrt(a^2 m^2+b^2)`. here `a^2=25 and b^2=16` substituting values we get=>` y=mx+-sqrt(25 m^2+16)` the point `(6,sqrt5)`, will satisfy this equation,so substituting values of x and y we get the eqn=> `11m^2 -12sqrt(5)m-11=0` using quadratic formula we get two values of m `m_1=(6sqrt(5)+sqrt(301))/(11) m_2=(6sqrt(5)-sqrt(301))/(11)` angle between the two tangents=`theta=tan^(-1)((m_1-m_2)/(1+m_1m_2))` substituting values of m_1 and m_2 we get=> `theta=.tan^(-1)oo=pi/2` |
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| 22. |
Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.`49 y^2-16 x^2=784` |
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Answer» Given equation is, `49y^2-16x^2 = 784` We will convert this equation into standard form. `=>49/784y^2-16/784x^2 = 1` `=> y^2/16-x^2/49 =1 ` So, this is our standard equation of hyperbola with, `a = 4, b = 7` `c = sqrt(a^2+b^2) = sqrt(16+49) = sqrt65` Here, as `y^2` is positive, major-axis will be `Y-`axis. Now, foci will be `(0,+-c) = (0,+-sqrt65).` Vertices will be `(0,+-a) = (0,+-4).` Eccentricity `= c/a = sqrt65/4.` Length of latus rectum ` = 2b^2/a = 2**49/4 = 49/2.` |
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| 23. |
Find the area of the region bounded by the latus recta of the ellipse `x^2/a^2+y^2/b^2=1` and the targets to the ellipse drawn at their ends |
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Answer» equation of tangent PR `(xx_1)/a^2+(yy_1)/b^2=1` putting the extreme values `(xe)/a+y/a`=1 -(1) equation of tangent QR `(xe)/a-y/b=1` -(2) adding 1 and 2 `(2xe)/a=2` `x=a/e` putting this value in equation 1 `(xe)/a+y/a=1` `a/e*e/a+y/a=1` y=0 we got point R(`a/e`,0) area of `triangle` PQR A=`1/2*PQ*FR` `A=1/2*(2b^2)/a*sqrt(a/e-ae)` `A=1/2*(2b^2)/a*sqrt(a(1/e-e)` `A=b^2/asqrt(a((1-e^2)/e)` `A=b^2/asqrt(b^2/(ae))` `A=b^2/asqrt(b^2/sqrt(a^2-b^2)` `a=b^3/(a(a^2-b^2)^(1/4)` |
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| 24. |
The ends of the major axis of an ellipse are (- 2, 4) and (2, 1). If the point (1, 3) lies on the ellipse,then find its latus rectum and eccentricity. |
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Answer» the major end points are (-2,4),(2,1) by distance formula `2a=sqrt(4^2+3^2)` `2a=sqrt(16+9)` `2a=5` `a=5/2` center of eclipse `((-2+2)/2,(4+1)/2)` equation of eclipse `(x-0)^2/a^2+(y-(5/2))^2/b^2=1` `x^2/(25/4)+(y-(5/2))^2/b^2=1` now, we have to find b putting the value(1,3) `4/25+1/4b_2=1` `1/4b^2=1-4/5=21/25` so `b^2=25/84` `x^2/(25/4)+(y-(5/2))^2/b62=1` length of latus rectum will be`=2b^2/a=(2*25*2)/(84*5)=5/21` eccentricity=`e=sqrt(1-b^2/a^2)=sqrt(1-(25/84)/(25/4)=sqrt(20/21)`. |
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| 25. |
If the eccentricity of an ellipse is `5/8`and the distance between its foci is `10 ,`then find the latusrectum ofthe ellipse. |
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Answer» Given that, eccentricity=`5/8i.e.,e=5/8` Let equation of the ellipse be `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` ltvbrgt since the foci of the ellipse is `)pmae,o)`. `therefore` Distance between foci =`sqrt((ae+ae)^(2))` `rArr 2sqrt(a^(2)e^(2))=10` [`because` distance between the foci=10] `rArr sqrt(a^(2)e^(2))=5` `rArr a^(2)e^(2)=25` `rArr a^(2)=(25xx64)/25` `thereforea=8` We know that, `rArr b^(2)=a^(2)(1-e^(2))` `rArr b^(2)=64(1-25/64)` `rArr b^(2)=64((64-25)/64)` `b^(2)=39` `therefore` Length of latusrectum of ellipse=`(2b^(2))/a=2(39/8)=39/4` |
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| 26. |
The line `3x-4y = 12` is a tangent to the ellipse with foci (-2, 3) and (-1, 0). Find the eccentricity of the ellipse. |
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Answer» `f1(-2,3) & f2(-1,0)` So, `2ae=sqrt((-2+1)^2+ (3-0)^2)=sqrt10``4a^2e^2=10` `a^2*(1-b^2/a^2)=5/2` So, `a^2-b^2=5/2` Equation of tangent,`y=(3x)/4-3` & `y=mx pmsqrt(a^2m^2 +b^2)` Hence, `m=3/4 & a^2(3/4)^2+ b^2=9`Therefore, `a^2(1+9/16)=9+5/2``a^2=186/25` Hence, `4*186/25*e^2=10` So, `e=sqrt(125/368)=(5sqrt(5))/(4sqrt(23)` |
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| 27. |
Find the range of eccentricity of the ellipse `x^2/a^2+y^2/b^2=1`, (where a > b) such that the line segment joining the foci does not subtend a right angle at any point on the ellipse. |
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Answer» we know `x^2y^2=a^2e^2` we need to prove b>ae `b^2>a^2e^2` `b^2>a^2-b^2` `2b^2>a^2` `b^2/a^2>1/2` `0`ein(0,1/sqrt2)` |
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| 28. |
Show that the set all points such that thedifference of their distances from `(4,0)a n d(-4,0)`is always equal to 2 represents a hyperbola. |
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Answer» Let the points be P (x,y). `therefore` Distances of P from (4,0)`sqrt((x+4)^(2))+y^^(2))` …(i) Now, `sqrt((x+4)^(2)+y^(2))-sqrt((x-4)^(2)+y^(2))=2` `sqrt((x+4)^(2)+y^(2))=2+sqrt((x-4)^(2)+y^(2))` On squaring both sides, we get `x^(2)+8x+16+y^(2)=4+x^(2)-8x+16+y^(2)+4sqrt((x-4)^(2)+y^(2))` `rArr16x-4=4sqrt((x-4)^(2)+y^(2))` `rArr4(4x-1)=4sqrt((x-4)^(2)+y^(2))` `rArr 16x^(2)-8x+1=x^(2)= 16-8x+y^(2)` `rArr 15x^(2)-y^(2)=15` which is parabola. |
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| 29. |
Show that the point `(x ,y)`given by `x=(2a t)/(1+t^2)a n dy=((1-t^2)/(1+t^2))`lies on a circle for all real values of `t`such that `-1lt=tlt=1,`where a is any given real number. |
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Answer» Given points are `x=(2at)/(1+t^(2))` and `y=(a(1-t^(2)))/(1+t^(2))` `because x^(2)+y^(2)=(4a^(2)t^(2))/((1+t^(2))^(2))+(a^(2)(1-t^(2)))/(1+t^(2))` `rArr1/(a^(2)(x^(2)+y^(2)))=(4t^(2)+1+t^(4)-2t^(2))/((1+t^(2))^(2))` `rArr1/(a^(2))(x^(2)+y^(2))=(t^(2)+2t^(2)+1)/((1+t^(2))^(2))` `rArr 1/(a^(2))(x^(2)+y^(2))=((1+t^(2))^(2))/((1+t^(2))^(2))` `rArr x^(2)+y^(2)=a^(2)`, which is a required circle. |
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| 30. |
Range of values of k for which the point (k,-1) is exterior to both the parabolas ` y^2 = |x|` is |
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Answer» `y^2=x` `y^2-x=0` 1-k>0 k<1 `y^2=-x` `y^2+x=0` `1+k>0` `k> -1` `k in (-1,1)` option 2 is correct. |
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| 31. |
The number of integer values of k for which the equation `x^2 +y^2+(k-1)x-ky+5=0` represents a circle whose radius cannot exceed 3 is |
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Answer» `x^2+y^2+2gx+2fy+c=0` where centre`=(-g,-f)` `r=sqrt(f^2+g^2-c)` `2g=k-1` `g=(k-1)/2` `2f=-k` `f=-k/2` `c=5` `r<=3` `sqrt(f^2+g^2-c) <=3` `f^2+g^2-c<=9` `k^2/4+(k-1)^2/4-5<=9` `(k^2+(k-1)^2-4xx14)/4<=0` `k^2+(k-1)^2-56<=0` `k^2+k^2-2k+1-56<=0` `2k^2-2k-55<=0` `k=2+-sqrt(4+440)/2xx2` `k=2+-sqrt2(111)/4` `k=1+-sqrt2(111)/2` `(k- (1-sqrt(111)/2)(k-(1+sqrt111/2)` `sqrt111=10.53` `(1-sqrt111)/2=-4.7` `(1+sqrt111)/2=5.7` option `1` |
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| 32. |
The equation of the image of the circle `x^2+y^2+16x-24y+183=0` by the line mirror `4x+7y+13=0` is : |
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Answer» `x^2+y^2+16x-24y+183=0` `x^2+16x+64-64+y^2-24y+144-144+183=0` `(x+8)^2+(y-12)^2+183-208=0` `(x+8)^2+(y-12)^2-25=5^2` Center(-8,12) `4x+7y+13=0-(1)` `7x+4y+k=0` `-56+48+k=0` `k=8` `7x+4y+8=0-(2)` From equation 1 and 2 `y=325/65=5` `y=5` `4x+7y+13=0` 9`4x+35+13=0` `x=-12` (-8,12),(-12,5),(h,k) `(-8+h)/2=-12` `h=-16` `(12+k)/2=5` `k=-2` (-16,-2) `(x+16)^2+(y+2)^2=25` `x^2+y^2+32x+4y+256+4-25=0` `x^2+y^2+32x+4y+235=0`. |
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| 33. |
A student is allowed to select at most n books from a collection of (2n+1) books. If the total number of ways in which a student selects at least one book is 63. then n equals to - |
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Answer» `(2n+1)C_1+(2n+1)C_2+...+(2n+1)C_n=63` `(2n+1)C_0+(2n+1)C_1+...+(2n+1)C_(2n+1)=2^(2n+1)` `(2n+1)C_0=(2n+1)C_(2n+1)=1` `(2n+1)C_1=(2n+1)C_(2n)` `1+1+2S=2^(2n+1)` `2+2*63=2^(2n+1)` `128=2^(2n+1)` `2^7=2^(2n+1) `2n+1=7` `2n=6` `n=3`. |
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| 34. |
Find the coordinates of points on the parabola `y^2=8x`whose focal distance is `4.` |
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Answer» Given parabola is `y^(2)`=8x On compairing this parabola to the `y^(2)`=4ax, we get…..(i) `8x=4axrArra=2` `therefore` Focal distance=`abs(x+a)=4` `rArr abs(x+2)=4` `rArr x+2=pm4` `rArr x=2,-6` But `xne-6` For x=2, `y^(2)=8xx2` `therefore y^(2)=16rArry=pm4` So, thepoints are (2,4) and (2,-4). |
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| 35. |
over the towers of a bridge a cable is hung in the form of a parabola, have their tops 30 meters above the road way are 200 meters apart. If the cable is 5 meters above the road way at the centre of the bridge, then the length of the vertical supporting cable 30 meters from the centre is |
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Answer» `x^2=4a(y-b)` `x^2=4a(y-5)` `100*100=4a(30-5)` a=100 `x^2=400(y-5)` At x=30 `900=400(y-5)` `y=29/4m` |
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| 36. |
Filnd the distance between the directricesthe ellipse `(x^2)/(36)+(y^2)/(20)=1.` |
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Answer» The equation of ellipse is `(x^(2))/36+(y^(2))/20=1` On comparing this equation with `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`we get `a=6,b=2sqrt5` we know that, `b^(2)=a^(2)(1-e^(2))` ltbr. `rArr 20=36(1-e^(2))` `rArr 20/36=1-e^(2)` `therefore e=sqrt(1-20/36)=sqrt(16/36)` `E=4/6=2/3` Now, directrices=`(+a/e,-1//e)` `therefore a/e=(6/2)/3=(6xx3)/2=9` and `-a/e=-9` `therefore` Distance between the directrices=`abs(9-(-9))=18` |
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| 37. |
The angle between the two tangents from the origin to the circle `(x-7)^2+ (y+1)^2= 25` equals |
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Answer» `sintheta=5/(5sqrt2)=1/sqrt2` `theta=pi/4` `OC=sqrt50=5sqrt2`. |
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| 38. |
Two parabola have the same focus. If their directrices are the x-axis and the y-axis respectively, then the slope of their common chord is : |
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Answer» F(n,k) D_1:y=0 D_2:x=0 `P_1:(x-h)^2+(y-k)^2=y^2` `sqrt((x-h)^2+(y-k)^2)=|(lx+my+n)/sqrt(l^2+m^2)|` `x^2-2hx+h^2+y^2-2yk+k^2=y^2` `P_1:x^2-2hx-2yk+h^2+k^2=0-(1)` `P_2:x^2-2hx+h^2+y^2-2yk+k^2=x^2` `y^2-2hx-2yk+h^2+k^2=0` Chord=`P_1=P_2` `x^2-2hx-2yk+h^2+k^2-y^2+2hx+2ykh^2-k^2=0` `y^2-x^2=0` `(y-x)(y+x)=0` `m=1,-1` option 1 is correct. |
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| 39. |
From the focus of parabola `y^2 = 8x` as centre, a circle is described so that a common chord is equidistant from vertex and focus of the parabola. The equation of the circle is |
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Answer» OP=PF OP=OF/2=1 `y^2=8x` `y^2=8` `y=pm2sqrt2` `FN=sqrt((2-1)^2+(2sqrt2-0)^2)=sqrt9=3` Circle=`(x-2)^2+y^2=9`. |
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| 40. |
The locus of the mid point of a chord of the circle `x^2+y^2=4` which subtends a right angle at the origin is |
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Answer» `/_OAD` and`/_OBD` OA=OB OD=OD AD=DB `/_OAD cong /_OBD` `/_AOD=/_BOD=90/2=45` `cos45=(OB)/(OA)` `1/sqrt2=sqrt(h^2+k^2)/2` `1/2=(h^2+k^2)/4` `h^2+k^2=2` `x^2+y^2=2` option 3 is correct. |
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| 41. |
P(-3, 2) is one end of focal chord PQ of the parabola `y^2+4x+4y=0`. Then the slope of the normal at Q is |
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Answer» Parabola=that tangent at one end of focal chord is parallel to normal at other end. Slope of normal at `theta`=slope of tangent at P `y^2+4x+4y=0` diff with respect to x `2ydy/dx+4+4dy/dx=0` `dy/dx=-4/(2y+4)` `=-1/2`. |
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| 42. |
Find the length of the normal chord which subtends an angle of `90^@` at the vertex of the parabola `y^2=4x` . |
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Answer» Let `PQ` is the common chord. Then, at any point coorinates of `P` and `Q` are `(t1^2,2t1)` and `(t2^2,2t2)` As the chord subtends an angle of `90^@`, relation between t1 and t2 will be, `t2=-t1-2/(t1)` `-> Eq(1)` If O is the origin, then OP and OQ are perpendicular to each other. In that case, both their slopes multiplication will be -1. Thus, `(2t1)/(t1^2)**(2t2)/(t2^2)=-1` `=>t1.t2=-4 ->Eq(2)` Putting values of t2 from Eq(1) `=>t1(-t1-2/(t1))=-4` `=>-t1^3-2t1=-4t1` `=>t1^3-2t1=0` Solving, above, we get, `t1=0` or `t1=sqrt(2)` As, t1=0 is not possiblee, so `t1 = sqrt(2)` Putting value of t1 in Eq(2), we get, `t2 = -2sqrt(2)` So, coordinates of chord are`P(2,2sqrt(2))` and `Q(8,-4sqrt(2))` So, length of PQ will be `sqrt((6sqrt(2))^2+6^2))= sqrt(108) = 6sqrt(3)` |
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| 43. |
In a circle of diameter 40cm the length of a chord is 20 cm Find the length of the minor arc |
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Answer» AB=40cm AC=r=20cm `/_PCQ=60^0` Minar arc=`2pi*20*60/360=20/3pi`. |
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| 44. |
The power of the origin w.r.t the circle on a focal chord of `y^2 = 4ax` as a diameter is |
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Answer» PQ is a diameter. `((at_1^2+at_2^2)/2,2a(t_1+t_2)/2)` `t_1t_2=-1` `(x-a)^2+y^2=4a^2` Power=`-a(3a)=-3a^2`. |
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| 45. |
If `y=2x` is the chord of the circle `x^2+y^2-4x=0`, find the equation of the circle with this chord as diameter. |
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Answer» `x^2+(2x)^2-4x=0` `x(5x-4)=0` x=0,4/5 `x_1=4/5,y_1=2x_1=8/5` `((x_1+0)/2,(y_1+0)/2)=(x_1/2,y_1/2)=(2/5,4/5)` `(x-h)^2+(y-k)^2=r^2` `(x-2/5)^2+(y-4/5)^2=(x_1/2)^2+(y_1/2)^2` `x^2-4/5x+-8y/5+y^2=0` `5(x^2+y^2)-4x-8y=0`. |
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| 46. |
If from any point `P` on the circle `x^2+y^2+2gx+2fy+c=0`, tangents are drawn to the circle `x^2+y^2+2gx+2fy+csin^2 alpha+(g^2+f^2)cos^2 alpha=0`, then the angle between the tangents is :(A) `alpha`(B) `2 alpha`(C) `alpha/2`(D) `alpha/3` |
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Answer» P(h,k) to the cords `x^2+y^2+2xy+2fy=0` `T=h^2+k^2+2gh+2fk+c` `T=h^2+k^2+2gh+2fk+sin^alpha+(g^2+f^2)cos^2alpha` `sintheta(h,k)` lies on the circle `x^2+y^2+2gx+2fy+c=0` `t^2=(h^2+k^2gh+2fk+c)+csin^alpha+(g^2+f^2)cos^2alpha-c` `=(g^2+f^2)cos^2alpha-c(1-sin^2alpha)` `r^2=g^2f^2-(csin^2alpha+(g^2+f^2)cos^2alpha)` `=(g^2+f^2-c)sin^2alpha` `tantheta=r/T=sqrt((g^2+f^2-c)sin^2alpha)/sqrt(g^2+f^2-c cos^2alpha)` `sqrt(sin^2alpha/cos^2alpha)=tanalpha` `theta=alpha` option b is coorect |
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| 47. |
The locus of the mid points of the chords of the circle `x^2+y^2+4x-6y-12=0` which subtend an angle of `pi/3`radians at its circumference is:(A) `(x-2)^2+(y+3)^2=6.25` (B) `(x+2)^2+(y-3)^2=6.25`(C) `(x+2)^2+(y-3)^2=18.75` (D) `(x+2)^2+(y+3)^2=18.75` |
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Answer» We can draw the figure from the given details. Please refer to the video for diagram. Equation of the given circle is is, `x^2+y^2+4x-6y-12 = 0` Comparing this equation with the standard circle equation, `x^2+y^2+2gx+2fy+c = 0`, we get `g = 2,f = -3, c = -12` Now, centre of circle will be `(-g,-f) = (-2,3)` And, radius AC will be `sqrt(g^2+f^2-c) = sqrt(4+9+12) = 5` Now, we are given ,`/_AQB = 60^@` `/_ACB` will be twice of it, as C is centre of the circle. `/_ACB = 2**60^@ = 120^@` Now, if `P(alpha,beta)` is the midpoint of the chord, then `/_ACP = 60^@` So, CP will be `/_ACcos60^@ = 5/2` Now, we can say that `CP^2 = ((alpha - (-2))^2 + (beta -3)^2)` `=>(alpha+2)^2+(beta-3)^2 = (2.5)^2` `=>(alpha+2)^2+(beta-3)^2 = 6.25` If we replace (alpha,beta) with (x,y), then, `(x+2)^2+(y-3)^2 = 6.25` that is the required equation. |
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| 48. |
The arithmetic mean of the ordinates of the feet of the normals from (2,6) to the parabola `y^2=4x` is |
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Answer» `(at^2, 2at) = (t^2,2t)` N`=> y= -tx + 2at + at^3` `y = -tx + 2t + t^3`-(2,6) `6 = -2t + 2t + t^3` `t^3 = 6` `t^3- 6=0` lets say , `t_1 + t_2 + t_3 = 0` mean =`(2at_1 + 2 at_2+ 2at_3)/2` `=0` Answer |
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| 49. |
If `a!=0` and the line `2bx+3cy+4d=0` passes through the points of intersection of the parabola `y^2 = 4ax` and `x^2 = 4ay`, then |
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Answer» Intersection points of `y^2 = 4ax` and `x^2 = 4ay` will be `(0,0)` and `(4a,4a)`. Putting these values of `(x,y)` in given line, `2bx+3cy+4d = 0` At point `(0,0),` `0+0+4d = 0 => d = 0->(1)` At point `(4a,4a),` ` 8ab+12ac+4ad = 0` `=>4a(2b+3c+0) = 0` `=>4a(2b+3c) = 0` As ,` a !=0,` So, `2b+3c = 0->(2)` Squaring and adding (1) and (2), `d^2+(2b+3c)^2 = 0` So, option `D` is the correct option. |
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| 50. |
Length of the latus rectum of the hyperbola `xy-3x-4y+8=0` |
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Answer» Equation of given hyperbola is, `xy-3x-4y+8 = 0` `=>x(y-3)-4(y-3)-12+8 = 0` `=>(x-4)(y-3) = 2^2` So, this is a standard form of rectangular hyperbola, `XY = a^2/2`. `:. a^2/2 = 2^2 = 4 => a^2 = 8=> a = 2sqrt2` Length of latus rectum of rectangular hyperbola `= 2a = 2**2sqrt2 = 4sqrt2` |
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