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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. |
Statement1: An equation of a common tangent to the parabola `y^2=16sqrt(3)x`and the ellipse `2x^2+""y^2=""4""i s""y""=""2x""+""2sqrt(3)`.Statement 2:If the line `y""=""m x""+(4sqrt(3))/m ,(m!=0)`is a common tangent to theparabola `y^2=""16sqrt(3)x`and the ellipse `2x^2+""y^2=""4`, then m satisfies `m^4+""2m^2=""24`.(1)Statement 1 isfalse, statement 2 is true(2)Statement 1 istrue, statement 2 is true; statement 2 is a correct explanation for statement1(3)Statement 1 istrue, statement 2 is true; statement 2 is not a correct explanation forstatement 1(4)Statement 1 istrue, statement 2 is false |
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Answer» `y^2 = 16 sqrtx` `2x^2 + y^2 = 4` so,`y= mx + 4sqrt3/m` `c^2 = a^2 m^2 + b^2` `((4 sqrt3)/m)^2 = 2m^2 + 4` `48/m^2 = 2m^2 + 4` `=> 24 = m^4 + 2m^2` 2)`y= 2x + 2sqrt3 ` `y= 2x + 4sqrt3/2` so, option 1 is correct |
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| 2. |
A normal drawn to the parabola `=4a x`meets the curve again at `Q`such that the angle subtended by `P Q`at the vertex is `90^0dot`Then the coordinates of `P`can be`(8a ,4sqrt(2)a)`(b) `(8a ,4a)``(2a ,-2sqrt(2)a)`(d) `(2a ,2sqrt(2)a)`A. `(8a,4sqrt(2)a)`B. (8a,4a)C. `(2a,-2sqrt(2)a)`D. `(2a,2sqrt(2)a)` |
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Answer» Correct Answer - C::D 3,4 `t_(2)=-t_(1)-(2)/(t_(1))` Also, `(2at_(1))/(at_(1)^(2))xx(2at_(2))/(at_(2)^(2))=-1` `ort_(1)t_(2)=-4` `:.(-4)/(t_(1))=-t_(1)-(2)/(t_(1))` `ort_(1)^(2)+2=4andt_(1)=pmsqrt(2)` So, the point can be `(2a,pm2sqrt(2)a)`. |
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| 3. |
Find the equation of the common tangent to the curves `y^2=8x` and xy=-1.A. 3y = 9x + 2B. y = 2x + 1C. 2y = x + 8D. y = x + 2 |
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Answer» Correct Answer - D Tangent to the curve `y^(2) = 8x` is `y = mx + (2)/(m)`. So it must satisfy xy = - 1 `implies x (mx + (2)/(m)) = - 1 implies mx^(2) + (2)/(m) x + 1 = 0` Since, it has equal roots. `:. D = 0` `implies (4)/(m^(2)) - 4m = 0` `implies m^(3) = 1` `implies m = 1` Hence, equation of common tangent is y = x + 2 |
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| 4. |
Maximum number of common normals of `y^2=4ax and x^2=4by` is ____A. 3B. 4C. 6D. 5 |
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Answer» Correct Answer - D (4) Normal to `y^(2)=4axandx^(2)=4by` in terms of m are `y=mx-2am-am^(3)` `andy=mx+2b+(b)/(m^(2))` For a common normal, `2b+(b)/(m^(2))+2am+am^(3)=0` `oram^(5)+2am^(3)+2bm^(2)+b=0` This means there can be at most five common normals |
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| 5. |
Find the equation of the common tangent to the curves `y^2=8x` and xy=-1.A. 3y=9y+2B. y=2x+1C. 2y=x+8D. y=x+2 |
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Answer» Correct Answer - D The equation of tangent to `y^(2)=8x` is `y=mx+2/m` If it touches xy=1, then `x(mx+2/m)=-1" must have equal roots"` `rArr" "mx^(2)+2/mx+1=0" must have equal roots"` `rArr" "4/m^(2)-4m=0rArrm^(3)-1=0rArrm=1.` Substituting the value of m in (i), we get y=x+2 as the equation of the common tangent. |
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| 6. |
If from a point A, two tangents are drawn to parabola `y^2 = 4ax` are normal to parabola `x^2 = 4by`, thenA. `a^(2)geb^(2)`B. `a^(2)ge4b^(2)`C. `a^(2)ge8b^(2)`D. `8a^(2)geb^(2)` |
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Answer» Correct Answer - C The equation of a tangent to `y^(2)=4ax` is `y=mx+a/mrArrx=y/m-a/m^(2)" ...(i)"` This will be a normal to the parabola `x^(2)=4by`, if it is of the form `x=y/m-(2b)/m-b/m^(3)" ...(ii)"` `:." "-a/m^(2)=-(ab)/m-b/m^(3)rArr2bm-am+b=0` Since m is real. `:." "a^(2)-8b^(2)ge0rArra^(2)ge8b^(2)` |
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| 7. |
The common tangent of the parabolas `y^(2)=4x" and "x^(2)=-8y,` isA. y=x+2B. y=x-2C. y=2x+3D. none of these |
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Answer» Correct Answer - D The equation tangent to `y^(2)=4x" is "y=mx + 1/m` If it touches `x^(2)=-8y` then the equation `x^(2)=-8(mx=1/m)` must have equal roots. `rArr" "64m^(3)=32rArr,(1/2)^("1/3")` Hence, the equation of the common tangent is `y=x/(2^("1/3"))+2^("1/3")` |
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| 8. |
If a, b, c are distinct positive real numbers such that the parabolas y2 = 4ax and y2 = 4c (x– b) will have a common normal, then(a) 0 < b/a - c < 1(b) b/a - c < 0(c) 1 <b/a - c < 2(d) b/a - c > 2 |
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Answer» Correct option (d) b/a - c > 2 Explanation: Equation of normals are y = mx –2am – am3 ......(1) y = m(x – b) – 2cm – cm3 ......(2) Equation 1 and 2 are identical then –2am – am3 = –bm –2cm –cm3 / m 2a + am2 = b + 2c + cm2 (a– c)m2 = b + 2(c – a) m2 = b/a - c - 2 m = ± √b/a - c - 2 For m be real b/a - c > 2 |
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| 9. |
Two equal parabolas have the same vertex and their axes are at right angles. The length of the common tangent to them, isA. 3aB. `3sqrt2a`C. 6aD. 2a |
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Answer» Correct Answer - B Let the two equal parabolas be `y^(2)=4ax` and `x^(2)=4ay.` The equation of any tangent to `y^(2)=4ax` is `y=mx+a/m" at the point"(a/m^(2),(2a)/m)` Now, `y=mx+a/mrArrx=y/m-a/m^(2)" ...(i)"` If if touches the parabola `x^(2)=4ay,` then `(-a)/m^(2)=a/(1//m)" "["Using : c"=a/m]` `(-a)/m^(2)=a/(1//m)" "["Using : c"=a/m]` `rArr" "m^(3)=-1rArr=-1` The common tangent touches `y^(2)=4ax" at "P(1/m^(2),(2a)/m)` So, the coordinates of the point of contact are `P(a, -2a)`. The common tangent (i) touches `x^(2)=4ay` at point Q whose coordinates are `((2a)/("1/m"),a/("a/m"^(2)))" "[becausex=my+a/m" touches "x^(2)=4ay" at "((2a)/m,a/m^(2))]` `=Q(2am, am^(2))=Q(-2a, a)" "[becausem=-1]` `:.` Length of the common tangent `=PQsqrt((a+2a)^(2)+(-2a-a)^(2))=3sqrt2a.` |
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| 10. |
From an external point `P ,`a pair of tangents is drawn to the parabola `y^2=4xdot`If `theta_1a n dtheta_2`are the inclinations of these tangents with the x-axis such that `theta_1+theta_2=pi/4`, then find the locus of `Pdot`A. `x - y +1 = 0`B. `x +y - 1 = 0`C. `x - y - 1 = 0`D. `x +y + 1 = 0` |
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Answer» Correct Answer - C Equation of tangent having slope m is `y = mx +(1)/(m)`, which passes through the point (h,k). `:. m^(2) h - mk +1 =0` `:. m_(1) + m_(2) = (k)/(h), m_(1)m_(2) = (1)/(h)` Given `theta_(1) + theta_(2) = (pi)/(4)` `rArr (m_(1)+m_(2))/(1-m_(2)m_(2)) =1` `rArr (k)/(h) =1 -(1)/(h)` `rArr y = x-1` |
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| 11. |
Find the points of contact `Q`and `R`of a tangent from the point `P(2,3)`on the parabola `y^2=4xdot` |
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Answer» Let points Q and R be `(t_(1)^(2),2t_(1))and(t_(2)^(2).2t_(2))`. Point of intersection of tangent at Q and R is `(t_(1)t_(2),(t_(1)+t_(2)))-=(2,3)`. `:." "t_(1)t_(2)=2andt_(1)+t_(2)=3`. Solving, we get, `t_(1)=1andt_(2)=2`. So, Q is (1,2) and R is (4,4). |
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| 12. |
If the parabolas `y^2=4a x`and `y^2=4c(x-b)`have a common normal other than the x-axis `(a , b , c`being distinct positive real numbers), then prove that `b/(a-c)> 2.`A. `0ltb/(a-c)lt1`B. `b/(a-c)gt2`C. `b/(a-c)lt0`D. `1ltb/(a-c)lt2` |
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Answer» Correct Answer - B The equation of any normal of slope m to the parabola `y^(2)=4c(x-c)` is `y=m(x-b)-2cm-cm^(2)or, y=mx-mb-2cm-cm^(3)` For this to be normal to `y^(2)=4ax`, we must have `-2am-am^(3)=-mb-2cm-cm^(3)` `rArr" "2a+am^(2)=b+2c+cm^(2)` `rArr" "m^(2)=(b+2c-2a)/(a-c)rArrm^(2)=b/(ac)-2rArrm=+-sqrt(b/(a-c)-2)` For m to be real, we must have `-b/(a-c)-2gt0rArr=b/(a-c)gt2` |
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| 13. |
Find the angle between the tangents drawn to `y^2=4x ,`where it is intersected by the line `y=x-1.` |
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Answer» Correct Answer - `90^(@)` The line y=x-1 passes through (1,0). That means it is a focal chord. Henec, the required angle is `90^(@)`. |
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| 14. |
Two tangent are drawn from the point `(-2,-1)`to parabola `y^2=4xdot`if `alpha`is the angle between these tangents, then find the value of `tanalphadot`A. 3B. 43468C. 2D. 43467 |
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Answer» Correct Answer - A The equation of a tangent to the parabola `y^(2)=4x` is `y=mx+1/m` If it passes through (-2, -1) then `-1=-2m+1/mrArr2m^(2)-m-1=0` Let `m_(1), m_(2)` be the roots of this equation. Then, `m_(1)+m_(2)=1//2" and "m_(1)m_(2)=-1//2` Now, `tanalpha+-(m_(1)-m_(2))/(1+m_(1)-m_(2))=+-(sqrt((m_(1)-m_(2))^(2)-4m_(1)-m_(2)))/(1+m_(1)-m_(2))` `rArr" "tanalpha=+-(sqrt(1//4+4//2))/(1-1//2)=+-(3//2)/(1//2)=+-3` |
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| 15. |
Find the angle between the tangents drawn from the origin to theparabolas `y^2=4a(x-a)` |
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Answer» Correct Answer - `90^(@)` The origin (0,0) lies on the directrix of the given parabola which is y=0. Then, the angle between the tangents is `90^(@)`. |
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| 16. |
If two tangents drawn from the point `(alpha,beta)`to the parabola `y^2=4x`are such that the slope of one tangent is double of the other, thenprove that `alpha=2/9beta^2dot` |
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Answer» Tangent to the parabola `y^(2)=4x` having slope m is `y=mx+(1)/(m)` It passes through `(alpha,beta)`. Therefore, `beta=malpha+(1)/(m)` `oralpham^(2)-betam+1=0` According to the question, it has roots `m_(1)and2m_(1)`. Now, `m_(1)+2m_(1)=(beta)/(alpha)andm_(1)*2m_(1)=(1)/(alpha)` `or2((beta)/(3alpha))^(2)=(1)/(alpha)` `oralpha=(2)/(9)beta^(2)` |
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| 17. |
Find the locus of the point of intersection of the perpendiculartangents of the curve `y^2+4y-6x-2=0`. |
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Answer» Correct Answer - 2x+5=0 We know that perpendicular tangents meet at the directrix. The given parabola is `y^(2)+4y-4x-2=0` `or" "(y+2)^(2)=6(x+1)` The equation of directrix is `x+1=-(6)/(4)orx=-(5)/(2)` `or" "2x+5=0` |
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| 18. |
The slope of normal to be parabola `y = (x^(2))/(4) -2` drawn through the point `(10,-1)` isA. `-2`B. `-sqrt(3)`C. `-1//2`D. `-5//3` |
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Answer» Correct Answer - C `x^(2) = 4(y+2)` is the given parabola. Any normal is `x = m (y+2) -2m -m^(3)` If `(10,-1)` lies on this line then `10 = m -2m -m^(3)` `rArrm^(3) + m + 10 =0 rArr m =-2` Slope of normal `= 1//m =- 1//2`. |
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| 19. |
Prove that the chord `y-xsqrt(2)+4asqrt(2)=0`is a normal chord of the parabola `y^2=4a x`. Also find the point on the parabola when the given chord is normal tothe parabola. |
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Answer» Correct Answer - `(2a,-2sqrt(2)a)` Equation of normal to the parabola `y^(2)=4ax` having slope m is `y=mx-2an-am^(3)` Comparing this equation with `y=sqrt(2)x-4asqrt(2)`, we get `m=sqrt(2)` `and" "2am+am^(3)=2sqrt(2)a+2sqrt(2)a=4asqrt(2)` Thus, given on the parabola is `(am^(2),-2am)-=(2a,-2sqrt(2)a)`. |
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| 20. |
Find the equation of normal to parabola `y=x^(2)-3x-4` (a) at point (3,-4) (b) having slope 5. |
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Answer» Correct Answer - (a) `3x-y-13=0`, (b) `5x-y-20=0` We have parabola `y=x^(2)-3x-4` Differentiating w.r.t. x, we get `(dy)/(dx)=2x-3` (a) `((dy)/(dx))_((2","-4))=3` So, equation of normal is `4+4=3(x-3)` `or" "3x-y-13=0` (b) `(dy)/(dx)=2x-3=5` `:." "x=4` So, y=16-12-4=0. Therefore, equation of normal is `y-0=5(x-4)` `or" "5x-y-20=0` |
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| 21. |
Three normals are drawn from the point (c, 0) to the curve `y^2 = x`. Show that c must be greater than 1/2. One normal is always the axis. Find c for which the other two normals are perpendicular to each other. |
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Answer» Correct Answer - `xgt(1)/(2),c=(3)/(4)` Equation of normal to parabola `y^(2)=4ax` having slope m is `y=mx-2am-am^(3)` For parabola `y^(2)=x,a=(1)/(4)`. Thus, equation of normal is `y=mx-(m)/(2)-(m^(3))/(4)` This, equation passes through (c,0). `:." "mc-(m)/(2)-(m^(2))/(4)=0` (1) `rArr" "m[c-(1)/(2)-(m^(2))/(4)]=0` `rArr" "m=0orm^(2)=4(c-(1)/(2))` (2) For, m=0 normal is y=0, i.e., x-axis. For other two normals to be real `m^(2)ge0` `rArr" "4(c-(1)/(2))ge0orcge(1)/(2)` For `c=(1)/(2),m=0`. Therefore, for other real normal, `cgt1//2`. Now, other two normals are perpendicular to each other. `:.` Product of their slope =-1 So, from (1), `4((1)/(2)-c)=-1` `rArr" "c=(3)/(4)` |
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| 22. |
If `y=2x+3`is a tangent to the parabola `y^2=24 x ,`then find its distance from the parallel normal. |
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Answer» Correct Answer - `(75)/(sqrt(5))` The equation of normal to the parabola `y^(2)=24x` having slope m is `y=mx-12m-6m^(3)`. It is parallel to y=2x+3. Therefore, m=2. Then the equation of the parallel normal is y=2x-24-48=2x-72 The distance between y=2x+3 and y=2x-72 is `|(72+3)/(sqrt(4+1))|=(75)/(sqrt(5))` |
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| 23. |
If the segment intercepted by the parabola `y=4a x`with the line `l x+m y+n=0`subtends a right angle at the vertex, then`4a l+n=0`(b) `4a l+4a m+n=0``4a m+n=0`(d) `a l+n=0`A. 4 al+ n =0B. 4al+4am +n =0C. 4am+n=0D. al+n=0 |
| Answer» Correct Answer - A | |
| 24. |
Find the locus of the midpoints of the portion of the normal to theparabola `y^2=4a x`intercepted between the curve and the axis. |
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Answer» Correct Answer - `y^(2)=a(x-a)` Normal at `P(at^(2),2at)" is "t=-tx+2at+at^(3)`. It meets the axis y=0 at `G(2a+at^(2),0)`. If (x,y) is the midpoint of PG, then `2x=2a+at^(2)+at^(2),2y=2at` Eliminating t, we have `x-a=at^(2)-a((y)/(a))^(2)` `or" "y^(2)=a(x-a)` which is the required locus. |
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| 25. |
If the line `x-1=0`is the directrix of the parabola `y^2-k x+8=0`, then one of the values of `k`is`1/8`(b) 8 (c) 4(d) `1/4`A. `1//8`B. 8C. 4D. 43469 |
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Answer» Correct Answer - C The equation of the directrix of this parabola is `x-8/k=-k/4" "["Using:"x=-a]` `rArr" "x=8/k-k/4` But, the equation of the directrix is given as x-1=0 `:." "8/k-k/4=1rArrk^(2)+4k-32=0rArrk=-8, 4.` |
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| 26. |
If the line `x-1=0`is the directrix of the parabola `y^2-k x+8=0`, then one of the values of `k`is`1/8`(b) 8 (c) 4(d) `1/4`A. -8B. `1//8`C. `1//4`D. 4 |
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Answer» Correct Answer - A::D 1,4 We have equation of parabola, `y^(2)=kx-8ory^(2)=k(x-(8)/(k))` Equation of directrix is `x-(8)/(k)=-(k)/(4)orx=-(k)/(4)+(8)/(k)` Given equation of directrix is x=1. `:.(8)/(k)-(k)/(4)=1` `rArrk^(2)+4k-32=0` `rArr(k-4)(k+8)=0` `rArrk=-8,4` |
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| 27. |
If the focus and vertex of a parabola are the points (0, 2) and (0, 4),respectively, then find the equation |
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Answer» Correct Answer - `x^(2)+8y=32` Focus is (0,2) and vertex is (0,4). so, axis of the parabola is y-axis. Also, parabola is concave downward. Distance between focus and directrix is a=2. `:.` Latus rectum = 4a=8 So, using equation `(x-h)^(2)=4a(y-k)`, equation of parabola is `(x-0)^(2)=-8(y-4)` `or" "x^(2)+8y=32` |
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| 28. |
PQ is a chord ofthe parabola `y^2=4x` whose perpendicular bisector meets the axis at M and the ordinate of the midpoint PQ meets the axis at N. Then the length MN is equal to |
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Answer» Correct Answer - 2 `P-=(t_(1)^(2),2t_(1))andQ-=(t_(2)^(2),2t_(2))` L is the midpoint of PQ. `:." "L-=((1)/(2)(t_(1)^(2)+t_(2)^(2)),(t_(1)+t^(2)))` `"Slope of PQ"=(2)/(t_(1)+t_(2))` `y-(t_(1)+t_(2))=(-(t_(1)+t_(2)))/(2)(x-(t_(1)^(2)+t_(2)^(2))/(2))` Putting y = 0, we get `M-=(2+(t_(1)^(2)+t_(2)^(2))/(2),0)` `N-=((t_(1)^(2)+t_(2)^(2))/(2),0)` `:." "MN=OM-ON=2` |
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| 29. |
The subtangent, ordinate and subnormal to the parabola `y^2 = 4ax` are inA. APB. GPC. HPD. none of these |
| Answer» Correct Answer - B | |
| 30. |
If focal distance of a point P on the parabola `y^(2)=4ax` whose abscissa is 5 10, then find the value of a. |
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Answer» Correct Answer - 5 Focal distance of point P is (a+x) or (a+5). Given that a+5=10 `:." "a=5` |
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| 31. |
The length of the subtangent to the parabola `y^(2)=16x` at the point whose abscissa is 4, isA. 2B. 4C. 8D. none of these |
| Answer» Correct Answer - C | |
| 32. |
At what point of the parabola x2 = 9y is the abscissa three times that of ordinate? |
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Answer» Given: The parabola, x2 = 9y Let us assume the point be (3y1, y1). Now by substituting the point in the parabola we get, (3y1)2 = 9(y1) 9y12 = 9y1 y12 – y1 = 0 y1(y1 – 1) = 0 y1 = 0 or y1 – 1 = 0 y1 = 0 or y1 = 1 The points is B (3(1), 1) => (3, 1) ∴The point is (3, 1). |
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| 33. |
Find the range of values of `lambda`for which the point `(lambda,-1)`is exterior to both the parabolas `y^2=|x|dot` |
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Answer» Correct Answer - `-1ltlamdalt1` The parabola are `y^(2)-x=0andy^(2)+x=0`. The point `(lamda,-1)` is an exterior point if `1-lamdagt0and1+lamdagt0` `or" "lamdaltandlamdagt-1` `or" "-1ltlamdalt1` |
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| 34. |
The polar of line point (2, 1) with respect to the parabola `y^(2)=6x,` isA. `y=3x+2`B. `y=3x+6`C. `3y=x+6`D. `y=3x+4` |
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Answer» Correct Answer - B The required polar is `y=6((x+2))/2rArry=3x+6` |
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| 35. |
Find the coordinates of any point on the parabola whose focus is (0, 1)and directrix is `x+2=0`A. `(t^(2)+1, 2t-1)`B. `(t^(2)+1, 2t+1)`C. `(t^(2), 2t)`D. `(t^(2)-1, 2t+1)` |
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Answer» Correct Answer - D The equation of the parabola is given by `sqrt((x-0)^(2)+(y-1)^(2))=|(x+2)/(sqrt1+0)|` `rArr" "x^(2)+(y=1)^(2)=(x+2)^(2)` `rArr" "(y-1)^(2)=4(x+1)` The parametrix coordinates of the point of this parabola are given by `x+1=t^(2)" and "y-1=2t" i.e. "(t^(2)-1, 2t+1)` |
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| 36. |
If the point `(2a,a)` lies inside the parabola `x^(2) -2x - 4y +3 = 0`, then a lies in the intervalA. `[(1)/(2),(3)/(2)]`B. `((1)/(2),(3)/(2))`C. `(1,3)`D. `((-3)/(2),(-1)/(2))` |
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Answer» Correct Answer - B The parabola is `(x-1)^(2) =4 (y-(1)/(2))` and origin lies outside the parabolic region, (0,0) makes `x^(2) - 2x - 4y +3` positive. `:. (2a,a)` should make `x^(2) -2x -4y +3` negative i.e., `4a^(2) - 8a + 3 lt 0` `rArr (2a-1) (2a-3) lt 0` Thus, a belongs to the open interval `((1)/(2),(3)/(2))`. |
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| 37. |
The focus of the parabola `x^2 -8x + 2y +7=0` isA. (4, 7/2)B. (4, 9/2)C. (9/2, 4)D. (1, 0) |
| Answer» Correct Answer - B | |
| 38. |
Find the coordinates of a point the parabola `y^(2)=8x` whose distance from the focus is 10. |
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Answer» Distance of point P(x,y) from the focus S(a,0) of the parabola `y^(2)=4ax` is SP = a+x. Also, given that SP=10. `:.` 10=2+x `:.` x=8 So, from `y^(2)=8x`, `y^(2)=64` `:. " "y=pm8` So, coordinates of point P are (8,8) or (8,-8). |
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| 39. |
Find the equation of the chord of the parabola `y^(2)=8x` having slope 2 if midpoint of the chord lies on the line x=4. |
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Answer» Let `M(4,y_(1))` be midpoint of the chord. So, equation of chord using equation `T=S_(1)` is `yy_(1)-4(x+4)=y_(1)^(2)-8xx4` `or" "yy_(1)-4x=y_(1)^(2)-16` Slope `=(4)/(y_(1))=2` (Given) `:." "y_(1)=2` So, equation of the chord is 2y-4x=-12 y=2x-6 |
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| 40. |
If `P S Q`is a focal chord of the parabola `y^2=8x`such that `S P=6`, then the length of `S Q`is6 (b) 4(c) 3 (d)none of theseA. 6B. 4C. 3D. none of these |
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Answer» Correct Answer - C (3) Since the semi-latus rectum of a parabola is the harmonic mean between the segments of any focal chord of the parabola, SP, 4,SQ are in H.P. Therefore, `4=(2SPxxSQ)/(SP+SQ)` `or4=(2(6)(SQ))/(6+SQ)` `or24+4(SQ)=12(SQ)` `orSQ=3` |
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| 41. |
If `P S Q`is a focal chord of the parabola `y^2=8x`such that `S P=6`, then the length of `S Q`is6 (b) 4(c) 3 (d)none of theseA. 6B. 4C. 3D. 8 |
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Answer» Correct Answer - C We have, `"Semi-latrusrectum = HM of SP and SQ"` `rArr" "4=(SPxxSQ)/(SPxxSQ)rArr4=(2xx6xxSQ)/(6+SQ)rArrSQ=3` |
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| 42. |
The normal at the point `P(ap^2, 2ap)` meets the parabola `y^2= 4ax` again at `Q(aq^2, 2aq)` such that the lines joining the origin to P and Q are at right angle. Then (A) `p^2=2` (B) `q^2=2` (C) `p=2q`(D) `q=2p`A. `p^(2)+pq+2=0`B. `p^(2)-pq+2=0`C. `q^(2)+pq+2=0`D. `p^(2)+pq+1=0` |
| Answer» Correct Answer - A | |
| 43. |
If `y_1, y_2, y_3`be the ordinates of a vertices of the triangleinscribed in a parabola `y^3=4a x ,`then show that the area of the triangle is `1/(8a)|(y_1-y_2)(y_2-y_3)(y_3-y_1)|dot`A. `1/(2a)|(y_(1)-y_(2))(y_(2)-y_(3))(y_(3)-y_(1))|`B. `1/(4a)|(y_(1)-y_(2))(y_(2)-y_(3))(y_(3)-y_(1))|`C. `1/(8a)|(y_(1)-y_(2))(y_(2)-y_(3))(y_(3)-y_(1))|`D. none of these |
| Answer» Correct Answer - C | |
| 44. |
If x-2y-a=0 is a chord of the parabola `y^(2)=4ax`, then its langth, isA. `4asqrt5`B. `40a`C. `20a`D. `15a` |
| Answer» Correct Answer - C | |
| 45. |
If `y_(1),y_(2),andy_(3)` are the ordinates of the vertices of a triangle inscribed in the parabola `y^(2)=4ax`, then its area isA. `(1)/(2a)|(y_(1)-y_(2))(y_(2)-y_(3))(y_(3)-y_(1))|`B. `(1)/(4a)|(y_(1)-y_(2))(y_(2)-y_(3))(y_(3)-y_(1))|`C. `(1)/(8a)|(y_(1)-y_(2))(y_(2)-y_(3))(y_(3)-y_(1))|`D. none of these |
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Answer» Correct Answer - C (3) Let `x_(1),x_(2),andx_(3)` be the abscissae of the points on the parabola whose ordinates are `y_(1),y_(2),andy_(3)`, respectively. Then `y_(1)^(2)=4ax_(1),y_(2)^(2)=4ax_(2),andy_(3)^(2)=4ax3`. Therefore, the area of the triangle whose vertices are `(x_(1),y_(1),(x_(2),y_(2)),and(x_(3),y_(3))` is `Delta=(1)/(2){:||(x_(1)" "y_(1)" "1),(x_(2)" "y_(2)" "1),(x_(3)" "y_(3)" "1)||=(1)/(2)||(y_(1)^(2)//4a" "y_(1)" "1),(y_(2)^(2)//4a" "y_(2)" "1),(y_(3)^(2)//4a" "y_(3)" "1)||=(1)/(8a)||(y_(1)^(2)" "y_(1)" "1),(y_(2)^(2)" "y_(2)" "1),(y_(3)^(2)" "y_(3)" "1)||` `=(1)/(8a)|(y_(1)-y_(2))(y_(2)-y_(3))(y_(3)-y_(1))|` |
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| 46. |
Let `A(x_(1), y_(1))" and "B(x_(2), y_(2))` be two points on the parabola `y^(2)=4ax`. If the circle with chord AB as a diameter touches the parabola, then `|y_(1)-y_(2)|=`A. 4aB. 8aC. `6sqrt2a`D. none of these |
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Answer» Correct Answer - B Let the coordinates of A and B `(at_(1)^(2), 2at_(2))" and " (at_(2)^(2), 2at_(2))` respectively. Then, equation of the circle described on AB as a diameter is `(x-at_(1)^(2))(x-at_(2)^(2))+(y-2at_(1))(y-2at_(2))=0` Suppose this cuts `y^(2)-4ax" at "(at^(2), 2at)`. Then, `(at^(2)-at_(1)^(2))(at^(2)-at_(2)^(2))+(2at+2at_(1))(2at-2at_(2))=0` `rArr" "(t+t_(1))(t+t_(2))+4=0` `rArr" "t^(2)+t(t_(1)+t_(2))+t_(1)t_(2)+4=0` If the circle touches the parabola, then this equation must give equal values of t. `:." "(t_(1)+t_(2))^(2)-4(t_(1)t_(2)+4)=0` `rArr" "(t_(1)-t_(2))^(2)=16rArr|t_(1)-t_(2)|=4rArr|y_(1)-y_(2)|=8a` |
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| 47. |
Let `A and B` be two distinct points on the parabola `y^2 = 4x`. If the axis of the parabola touches a circle of radius `r` having `AB` as its diameter, then the slope of the line joining `A and B` can be (A) `- 1/r` (B) `1/r` (C) `2/r` (D) `- 2/r`A. `+-1/r`B. `+-2/r`C. `+-3/r`D. `+-1/2r` |
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Answer» Correct Answer - B Let `A(t_(1)^(2), 2t_(1))" and "B(t_(2)^(2), 2t_(2))` be two point on `y^(2)=4x`. The coordinates of the cetre of the circle are `((t_(1)^(2)+t_(2)^(2))/2,t_(1)+t_(2))` Since the circle with diameter AB touches the axis of the parabola and is of radius r. `:. "t_(1)+t_(2)=+r` `:." Slope of AB"2/(t_(2)+t_(1))=+-2/r` |
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| 48. |
Find the equation of the parabola with focus F(4, 0) and directrix x=-4. |
| Answer» Correct Answer - `y^(2)=16x` | |
| 49. |
Find the equation of the parabola with vertex at the origin and focus at F(-2, 0). |
| Answer» Correct Answer - `y^(2)-8x` | |
| 50. |
Find the shortest distance between the line `x - y +1 = 0` and the curve `y^2 = x.`A. `sqrt3/4`B. `(3sqrt2)/8`C. `8/(3sqrt2)`D. `4/sqrt3` |
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Answer» Correct Answer - B Let `P(t^(2), t)` be point on the curve `x-=y^(2)` and S be the distance between P and the line `y-x-1=0`. Then, `S=|(t-t^(2)-1)/(sqrt(1+1))|=(t^(2)-t+1)/(sqrt2)=1/sqrt2{(t-1/2)^(2)+(sqrt3/2)^(2)}` Clearly, S is minimum when`t=1/2` For this value of t, we get |
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