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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.
| 101. |
Afocus of an ellipse is at the origin. The directrix is the line `x""=""4`and the eccentricity is 1/2. Then the lengthof the semimajor axis is(1) `8/3`(2)`2/3`(3)`4/3`(4)`5/3` |
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Answer» `a/e - ae = 4` `a(1- e^2)/e = 4` `a((1-1/4)/(1/2)) = 4` `a*(3/4)/(1/2) = 4` `a = 4*1/2 *4/3 ` `a= 8/3` option 1 is correct |
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| 102. |
An ellipse has eccentricity `1/2` and one focus at the point `P(1/2,1)`. Its one directrix is the comionand tangent nearer to the point the P to the hyperbolaof `x^2-y^2=1` and the circle `x^2+y^2=1`.Find the equation of the ellipse. |
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Answer» Correct Answer - A::B::C there are two common tangent to the circle `x^(2)+y^(2)=1` and the hyperbola `x^(2)-y^(2)=1` these are x=1 and x=-1. But x=1 is nearer to the point P(1/2,1). Now, if Q(x,y) is any point on the eillpse then its distance from the foruc is `QP= sqrt((x-1//2)^(2)+(y-1)^(2))` and its distance from the directrix is |x-1|. By definition of ellipse , `QP=e|x-1|implies sqrt((x-(1)/(2))^(2)+(y+1)^(2))=(1)/(2)|x-1|` `implies (x-(1)/(2))^(2)+(y-1)^(2)=(1)/(4)(x-1)^(2)` `implies x^(2)-x+(1)/(4)+y^(2)-2y+1=(1)/(4)(x^(2)-2x+1)` `implies 4x^(2)-4x+1+4y^(2)-8y+4=x^(2)-2x+1` `implies 3x^(2)-2x+4y^(2)-8y+4=0` `3[(x-(1)/(3))^(2)-(1)/(9)]+4(y-1)^(2)=0` `implies 3(x-(1)/(3))^(2)+4(y-1)^(2)=(1)/(3)` `implies ((x-(1)/(3)^(2)))/(1//9)`+((y-1)^(2))/(1//12)=1` |
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| 103. |
Find the equation of ellipse having focus at (1,2) corresponding directirx `x-y=2=0` and eccentricity `0.5`. |
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Answer» Correct Answer - `7x^(2)+7y^(2)+2xy-20x-28y+36=0` Equation of ellipse is locus of point (x,) such that SP= e. PM, where `S-=(1,2)` and M is foot of perpendicular from focus upon directirx. Therfore, the equation of ellipse is `sqrt((x-1)^(2)+(y-2)^(2))=(1)/(2)(|x-y+2|)/(sqrt(1^(2)+(-1)^(2)))` or `8(x^(2)-2x+1-y^(2)-4y+4)=x^(2)+y^(2)+4+4x-4y-2xy` or `7x^(2)+7y^(2)+2xy-20y+36=0` |
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| 104. |
Find the eccentricity, one of the foci, thedirectrix, and the length of the latus rectum for the conic `(3x-12)^2+(3y+15)^2=((3x-4y+5)^2)/(25)`. |
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Answer» Correct Answer - e= 1/3; focus `-=` (4,5); directrix : 3x-yy+5=0; L.R.=2/5 `(3x-12)^(2)+(3y+15)^(2)=((3x+4y+5)^(2))/(25)` or `sqrt((x-4)^(2)+(y+5)^(2))=(1)/(3)(|3x-4y+5|)/(5)` Hence the ratio of distance of he variable point P(x,y) from the foxed point (focus) (4,5) to that from the fixed line (directrix`-=3x-4y+5=0)` is 1/3 Also, the locus is an ellipse and its eccentricity is 1/3 Alos, lenght of the latus reactum `=2(e)xx` (Distance of (4,5) from the line 3x-4y+5=0 `=2xx(1)/(3)(|3xx4-4xx5+5|)/(5)=(2)/(5)` |
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| 105. |
A cruve is respresented by `C=21x^(2)-6xy+29y^(2)+6x-58y-151=0` The eccentricity of the cruve isA. `1//3`B. `1//sqrt(3)`C. `2//3`D. `2//sqrt(5)` |
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Answer» `24x^(2)-6xy+29y+6x-58y-151=0` `2(x-3y+3)^(2)+2(3x+y-1)^(2)=180` or `((x-3y+3)^(2))/(60)+((3x+y-1)^(2))/(90)=1` or `((x-3y+3)/(sqrt(1+3^(2))sqrt(6)))^(2)+((3x+y-1)/(3sqrt(1+3^(2))))=1` Thus, C is an ellipse whose lengths of axes are `6,2sqrt(6)`. The minor and the major axes are `x-3y+3=0 and 3x+y-1=0`, respectively. Their point of intersection gives the center of the center of the conic. Therefore, Center `-=(o,1)` |
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| 106. |
A cruve is respresented by `C=21x^(2)-6xy+29y^(2)+6x-58y-151=0` The lengths of axes areA. `6,2 sqrt(6)`B. `5,2sqrt(5)`C. `4,4sqrt(5)`D. none of these |
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Answer» `24x^(2)-6xy+29y+6x-58y-151=0` `2(x-3y+3)^(2)+2(3x+y-1)^(2)=180` or `((x-3y+3)^(2))/(60)+((3x+y-1)^(2))/(90)=1` or `((x-3y+3)/(sqrt(1+3^(2))sqrt(6)))^(2)+((3x+y-1)/(3sqrt(1+3^(2))))=1` Thus, C is an ellipse whose lengths of axes are `6,2sqrt(6)`. The minor and the major axes are `x-3y+3=0 and 3x+y-1=0`, respectively. Their point of intersection gives the center of the center of the conic. Therefore, Center `-=(o,1)` |
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| 107. |
A cruve is respresented by `C=21x^(2)-6xy+29y^(2)+6x-58y-151=0` The center of the conic C isA. (1,0)B. (0,0)C. (0,1)D. none of these |
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Answer» `24x^(2)-6xy+29y+6x-58y-151=0` `2(x-3y+3)^(2)+2(3x+y-1)^(2)=180` or `((x-3y+3)^(2))/(60)+((3x+y-1)^(2))/(90)=1` or `((x-3y+3)/(sqrt(1+3^(2))sqrt(6)))^(2)+((3x+y-1)/(3sqrt(1+3^(2))))=1` Thus, C is an ellipse whose lengths of axes are `6,2sqrt(6)`. The minor and the major axes are `x-3y+3=0 and 3x+y-1=0`, respectively. Their point of intersection gives the center of the center of the conic. Therefore, Center `-=(o,1)` |
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| 108. |
Find the equation of the set of all points whosedistances from (0,4) are `2/3`of their distances from the line `y=9.` |
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Answer» `(sqrt((h-0)^2+(k-4)^2))^2=(2/3|(k-9)/sqrt(1^2)|)^2` taking root both side `h^2+(k-4)^2=4/9((k-9)^2/1)` `9h^2=9(k-4)^2=4(k-9)^2` `9h^2+9k^2+144-72k=4k^2+324-72k` `9h^2+9k^2-4k^2+144-324=0` `9h^2+5k^2=180` `(9h^2)/180+(5k^2)/180=1` `h^2/20+k^2/36=1` putting x=h,y=k `x^2/20+y^2/36=1`. |
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| 109. |
A point moves so that its distance from the point (2,0) is always (1)/(3)` of its distances from the line x-18. If the locus of the points is a conic, then length of its latus rectum isA. `(16)/(3)`B. `(32)/(3)`C. `(8)/(3)`D. `(15)/(4)` |
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Answer» Clearly, locus is ellipse with eccetntircity `e=(1)/(3)`. Hence, focus is (2,0) and directrixd is x-18=0 Distance of focus from directrix=16 `rArra//e-ae=16` `:.(8)/(3)a=16` `:.a=6` `:. b^(2)=a^(2)(1-e^(2))=36(1-1//9)=32` `:. L.R.=(2b^(2))/(A)=(32)/(3)` |
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| 110. |
A point moves so that the sum of the squares ofits distances from two intersecting straight lines is constant. Prove thatits locus is an ellipse. |
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Answer» OA and DB `y=ntantheta,y=-xtantheta` `xsintheta-ycostheta=0` `xsintheta+ycostheta=0` `PL^2+PM^2=lambda^2`(const) `((hsintheta-kcostheta)/sqrt(sin^2theta+cos^2theta))^2+((hsintheta+kcostheta)/sqrt(sin^2theta+cos^2theta))^2=lambda^2` `h^2sin^2theta+k^2cos^2theta=lambda^2` `h^2/((lambda^2)/(sin^2theta))+k^2/((lambda^2/cos^2theta))=1` `(x^2)/(lambda/sintheta)^2+k^2/(lambda/costheta)^2=1`. |
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| 111. |
There are exactly two points on the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1`whose distances from its center are the same and are equal to `(sqrt(a^2+2b^2))/2dot`Then the eccentricity of the ellipse is`1/2`(b) `1/(sqrt(2))`(c) `1/3`(d) `1/(3sqrt(2))`A. `1//2`B. `1//sqrt(2)`C. `1//3`D. `1//3sqrt(2)` |
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Answer» (2) Since there are axactly two points on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` whosse distance from the center is the same, the poitns would either be the endpoints of the major axis or of the minor axis. But `sqrt((a^(2)+2b^(2)))//bgtb` So, the points are the vertices of the major axis hence, (1)`a=sqrt((a^(2)+2b^(2))/(2))` or `a^(2)=2b^(2)` `or e=sqrt(1-(b^(2))/(a^(2)))=(1)/(sqrt(2))` |
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| 112. |
If the maximum distance of any point on the ellipse `x^2+2y^2+2x y=1`from its center is `r ,`then `r`is equal to`3+sqrt(3)`(b) `2+sqrt(2)``(sqrt(2))/(sqrt(3-sqrt(5)))`(d) `sqrt(2-sqrt(2))`A. `3+sqrt(3)`B. `2+sqrt(2)`C. `sqrt(2)//sqrt(3-sqrt5)`D. `sqrt(2-sqrt(2))` |
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Answer» Herem the center of the ellipse is (0,0) Let `P(r cos theta, r sin theta)` be any point on the given ellipse. Then `r^(2) cos^(2) theta+2r^(2) sin ^(2) theta+2r^(2) sin theta cos theta=1` `or r^(2)=(1)/(cos^(2)theta+2 sin^(2) theta+sin 2 theta)` `or (1)/(sin^(2)theta+1sin 2theta)` `=(2)/(1-cos 2 theta+2+2sin 2 theta)` `(2)/(3-cos 2 theta+2 sin2 theta)` `or r_("max")=(sqrt(2))/(sqrt(3-sqrt(5)))` |
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| 113. |
If the curves `(x^2)/4+y^2=1`and `(x^2)/(a^2)+y^2=1`for a suitable value of `a`cut on four concyclic points, the equation of the circle passingthrough these four points is`x^2+y^2=2`(b) `x^2+y^2=1``x^2+y^2=4`(d) none of theseA. `x^(2)+y^(2)=2`B. `x^(2)+y^(2)=1`C. `x^(2)+y^(2)=4`D. none of these |
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Answer» The equation of conic through the point of intersection of given two ellipse is `((x^(2))/(4)+y^(2)-1)+lambda((x^(2))/(a^(2))+y^(2)-1)=0` or `x^(2)((1)/(4)+(lambda)/(a^(2)))+y^(2)(1+lamda)=1+lambda` or `x^(2)((a^(2))/(4a^(2)(1+lambda)))+y^(2)=1` Therefore , the circle is `x^(2)+y^(2)=1` |
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| 114. |
A point on the ellipse `x^2+3y^2=37`where the normal is parallel to the line `6x-5y=2`is`(5,-2)`(b) (5, 2)(c) `(-5,2)`(d) `(-5,-2)`A. (5,-2)B. (5,2)C. (-5,2)D. (-5,-2) |
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Answer» Differnetiating the equation of ellipse, `x^(2)+3y^(2)=37` w.r.t., x we get `(dy)/(dx)=-(x)/(3y)` The slop of the given line is 6/5 , which is normal to the ellipse Hence, `3x//y=6//5 or 2x=5y` Points in options (2) and (4) are satisfying the above equation adn that of ellipse. |
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| 115. |
The length of major ofthe ellipse `(5x-10)^2 +(5y+15)^2 = 1/4(3x-4y+7)^2` isA. 10B. `20//3`C. `20//7`D. 4 |
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Answer» `(5x-10)^(2)+(5y+15)^(2)=((3x-4y+7)^(2))/(4)` `or (x-2)^(2)+(y+3)^(2)=((1)/(2)(3x-4y+7)/(5))^(2)` `or sqrt((x-2)^(2)+(y-3)^(2))=(1)/(2)(|3x-4y+7|)/(5)` It is an ellipse , whose focus is (2,-3) directrix is 3x-4y+7=0, and eccentricity is 1/2. Length of perpendicular from the focus to the drectrix is `(|3xx2-4(-3)+7|)/(5)=5` or `(a)/(e)-ae=5` `or 2a-(a)/(2)=5` `or a=(10)/(3)` So, the length of the major of axis is 20/3 |
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| 116. |
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.`(x^2)/4+(y^2)/(25)=1` |
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Answer» Correct Answer - (i) 10 units, 4 units (ii) `A(0, -5) and B(0, 5)` (iii) `F_(1)(0, -sqrt(21)) and F_(2) (0, sqrt(21))" "` (iv) `e=sqrt(21)/5 " "` (v) ` 8/5 ` units |
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| 117. |
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.`(x^2)/(16)+(y^2)/9=1` |
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Answer» Correct Answer - (i) 8 units, 6 units (ii) `A(0, -4) and B((0, 4)` (iii) `F_(1)(0, -sqrt7) and F_(2) (0, sqrt7)" "` (iv) ` e= sqrt7/4" "` (v) ` 4 1/2` units |
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| 118. |
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.`(x^2)/(49)+(y^2)/(36)=1` |
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Answer» Correct Answer - (i) 14 units, 12 units (ii) `A (-7, 0) and B(7,0) ` (iii) `F_(1)(-sqrt(13),0) and F_(2) (sqrt(13), 0)" "` (iv) ` e = sqrt(13)/7" "` (v) `72/7 ` units |
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| 119. |
Let d be the perpendicular distance from the centre of the ellipse `x^2/a^2+y^2/b^2=1` to the tangent drawn at a point P on the ellipse. If `F_1 & F_2` are the two foci of the ellipse, then show the `(PF_1-PF_2)^2=4a^2[1-b^2/d^2]`. |
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Answer» The equation of the tangent at the point `P(cos theta, b sin theta)` on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` is `(x)/(a) cos theta+(y)/(b) sin theta=1" "(1)` The perpendicualr distance of (1) from the centre (0,0) of the ellipse is given by `d=(1)/(sqrt((1)/(a^(2))cos^(2)theta+(1)/(b^(2))cos^(2)theta))=(ab)/(sqrt(b^(2)cos^(2)theta+a^(2)sin^(2)theta))` `:. 4a^(2)(1-(b^(2))/(d^(2)))=4a^(2){1-(b^(2)cos^(2)theta+a^(2)sin^(2)theta)/(a^(2))}` `4(a^(2)-b^(2))cos^(2)theta=4a^(2)e^(2)cos^(2)theta" "(2)` The foci are `F_(1)-=(ae,0)and F_(2)-=(-ae),o)`. Therefore, `PF_(1)=(1-ecostheta)` and `PF_(2)=a(1+ e cos theta)` `:. (PF_(1)-PF_(2))^(2)=4a^(2)e^(2)cos^(2)theta" "(3)` Hence, from (2) and (3), we get have `(PF_(1)-PF_(2))^(2)=4a^(2)(1-(b^(2))/(d^(2)))` |
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| 120. |
If the midpoint of the chord of the ellipse `x^2/16+y^2/25=1`is `(0,3)` |
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Answer» The equation of the chord chose midpoind is (0,3) is `(3y)/(25)-1=(9)/(25)-1` i.e., y=3 It intersects th ellipse `(x^(2))/(16)=(y^(2))/(25)=1` `:. (x^(2))/(16)=1-(9)/(25)=(16)/(25) or x=+-(16)/(3)` `:.` Length of chord `=(32)/(5)` |
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| 121. |
Find the set of those value(s) of `alpha` for which `(7-(5alpha)/4,alpha)` lies inside the ellipse `x^2/25+y^2/16=1` |
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Answer» Correct Answer - B `(7-(5)/(4)alpha, alpha)` lies inside the ellipse `(x^(2))/(25) + (y^(2))/(16) =1` `rArr ((7-(5)/(4)alpha)^(2))/(25)+(alpha^(2))/(16)-1 lt 0` `rArr ((28-5 alpha)^(2))/(400) + (alpha^(2))/(16) -1 lt 0` `rArr (28 - 5 alpha)^(2) + 25 alpha^(2) - 400 lt 0` `rArr 50 alpha^(2) - 140 alpha - 192 lt 0` `rArr (5 alpha -12) (5 alpha -16) lt 0` `rArr (12)/(5) lt alpha lt (16)/(5)` |
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| 122. |
The ratio of the area enclosed by the locus of the midpoint of PS and area of the ellipse is (P-be any point on the ellipse and S, its focus)A. `(1)/(2)`B. `(1)/(3)`C. `(1)/(5)`D. `(1)/(4)` |
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Answer» Correct Answer - D Let ellipse be `(x^(2))/(a^(2)) +(y^(2))/(b^(2)) =1` Mid point of PS is (h,k) `:. h = (a cos theta +ae)/(2) rArr cos theta = (2h-ae)/(a)` and `k = (b sin theta)/(2)` Eliminating `theta`, we get locus as `((2h-ae)^(2))/(a^(2)) +(4k^(2))/(b^(2)) =1` `rArr ((x-(ae)/(2)))/((a^(2))/(4)) +(y^(2))/((b^(2))/(4))=1`, which is ellipse having enclosed area Area `rArr pi(a)/(2).(b)/(2) = (piab)/(4). :.` ratio `= (1)/(4)` |
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| 123. |
Find the maximum length of chord of the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` then find the locus of midpoint of PQ |
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Answer» Correct Answer - 4 Let `P-=(2sqrt(3)cos theta, 2 sin theta)` and `Q-=(2sqrt(3)cos((pi)/(2)+theta),2sin((pi)/(2)+theta))` or `Q-=(-2sqrt(2)sin theta, 2 cos theta)` `(PQ)^(2)=8(cos theta+sin theta)^(2)+4(sin theta-cos theta)^(2)=12+4 sin 2 theta` `:. (PQ)_("max")=4` |
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| 124. |
The locus of mid-points of a focal chord of the ellipse `x^2/a^2+y^2/b^2=1` |
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Answer» Let the midpoint of the focal chord the given ellipse be (h,k) . Then its equation is `(hx)/(k)+(ky)/(b^(2))=(h^(2))/(a^(2))+(k^(2))/(b^(2))" " ["Using"T=S_(1)]` Since this passes thoughb (ae,0), we have or `(hae)/(a)=(h^(2))/(a^(2))+(k^(2))/(b^(2))` Therefore, the locus of (h,k) is `(ex)/(a)=(x^(2))/(a^(2))+(y^(2))/(b^(2))` |
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| 125. |
For all real p, the line `2px+ysqrt(1-p^(2))=1` touches a fixed ellipse whose axex are the coordinate axes The locus of the point of intersection of perpendicular tangent isA. `x^(2)+y^(2)=5//4`B. `x^(2)+y^(2)=3//2`C. `x^(2)+y^(2)=2`D. none of these |
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Answer» Let the ellipse be `(x^(2))/(y^(2))+(y^(2))/(b^(2))=1` The line `y==mx+-sqrt(a^(2)m^(2)+b^(2))` touches, the ellipse for all m. Hence, it is identical with `y=-(2x)/(sqrt(1-p^(2)))+(1)/(sqrt(1-p^(2)))` Hence, `m=-(2p)/(sqrt(1-p^(2)))` and `a^(2)m^(2)+b^(2)=(1)/(1-p^(2))` or `a^(2)(4p^(2))/(a-p^(2))+b^(2)(1)/(1-p^(2))` or `p^(2)(4a^(2)-b^(2))+b^(2)-1=0` This equation is true for all real p if `b^(2)=1 and 4a^(2)=b^(2)` `b^(2)=1 and a^(2)=(1)/(4)` Therefore, the equation of the ellipse is `(x^(2))/(1//4)+(y^(2))/(1)=1` If e is its eccentricity, then `(1)/(4)=1-e^(2) or e^(2)=(3)/(4) or e=(sqrt(3))/(2)` be `=(sqrt(3))/(2)` Hence, the foci are `(0,+-sqrt(3)//2)` The equation of director circle is `x^(2)+y^(2)=(5)/(4)` |
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| 126. |
The locus of the point of intersection of the tangent at the endpointsof the focal chord of the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1(bA. circleB. ellipseC. hyperbolaD. pair of straight lines |
| Answer» Since the locus of the point of intersection of the tangent at the endpoints of a focal is directirx, the requied locus is `x=+-a//e` , which is pair of straight lines. | |
| 127. |
The equation of the line passing through the center and bisecting thechord `7x+y-1=0`of the ellipse `(x^2)/1+(y^2)/7=1`isA. x=yB. 2x=yC. x=2yD. x+y=0 |
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Answer» Let (h,k) be the midpoint of the chord 7x+y-1=0. Then, `(hx)/(1)+(ky)/(7)=(h^(2))/(1)+(k^(2))/(7)" "(1)` and `7x+y=1 " "(2)` represent the same straight line. Therefore, `(h)/(7)=(k)/(7) or h=k` Thus, the equation of the line joining (0,0) adn (h,k) is ty-x=0 |
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| 128. |
Point P represent the complex number z=x+iy and point Qrepresents the complex number z+1/z. If P moves on the circle |z|=2, then the eccentricity of locus of point Q isA. `3//5`B. `4//5`C. `3//4`D. `1//2` |
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Answer» Let `Q-=alpha+ibeta` Given that |z|=2, where z=x+iy `:. x^(2)+y^(2)=4` Now, `alpha+ibeta=z+(1)/(z)=(x+iy)+(1)/(x+iy)` `=(x+iy)+((x-iy)/(4))=(5x)/(4)+(3iy)/(4)` `:. alpha=(5x)/(4)and beta=(3y)/(4)` Since `x^(2)+y^(2)=4` `(16alpha^(2))/(25)+(16beta^(2))/(9)=4` ltbr So, locus of point Q is `(x^(2))/(25)+(y^(2))/(9)=(1)/(4)` Eccentricity of theis conic is given by `e^(2)=1-(9)/(25)=(16)/(25)` |
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