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Parabola question

2020 · Shift 1 · Q22
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  5. /2020 · Shift 1 · Q22

Parabola question

2020 · Shift 1 · Q22

JEE AdvancedMathematicsParabolaMCQ+3 / −1
Let a, b and λ\lambdaλ be positive real numbers. Suppose P is an end point of the latus return of the parabola y2 = 4 λ\lambdaλ x, and suppose the ellipse x2a2+y2b2=1{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1a2x2​+b2y2​=1 passes through the point P. If the tangents to the parabola and the ellipse at the point P are perpendicular to each other, then the eccentricity of the ellipse is
  1. A
    12{1 \over {\sqrt 2 }}2​1​
  2. B
    12{{1 \over 2}}21​
  3. C
    13{{1 \over 3}}31​
  4. D
    25{{2 \over 5}}52​
View written solutionFree

Correct answer: A

  1. Find the endpoint of the latus rectum of the parabola

For the parabola y2=4λx,y^2 = 4\lambda x,y2=4λx, its latus rectum is the line through the focus perpendicular to the axis, i.e. x=λx=\lambdax=λ.

Substituting x=λx=\lambdax=λ into the parabola: y2=4λ(λ)=4λ2  ⟹  y=±2λ.y^2 = 4\lambda(\lambda)=4\lambda^2 \implies y=\pm 2\lambda.y2=4λ(λ)=4λ2⟹y=±2λ.

So the endpoints of the latus rectum are P=(λ,±2λ).P=(\lambda,\pm 2\lambda).P=(λ,±2λ).

We can take P=(λ,2λ)P=(\lambda,2\lambda)P=(λ,2λ) because the ellipse and slope condition will give the same result for −2λ-2\lambda−2λ.

  1. Use the condition that the ellipse passes through PPP

The ellipse is x2a2+y2b2=1.\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.a2x2​+b2y2​=1. Since P=(λ,2λ)P=(\lambda,2\lambda)P=(λ,2λ) lies on it, λ2a2+4λ2b2=1.\frac{\lambda^2}{a^2}+\frac{4\lambda^2}{b^2}=1.a2λ2​+b24λ2​=1. So, λ2(1a2+4b2)=1.(1)\lambda^2\left(\frac1{a^2}+\frac4{b^2}\right)=1. \qquad (1)λ2(a21​+b24​)=1.(1)

  1. Find the tangent slope to the parabola at PPP

Differentiate y2=4λxy^2=4\lambda xy2=4λx implicitly: 2ydydx=4λ2y\frac{dy}{dx}=4\lambda2ydxdy​=4λ so dydx=2λy.\frac{dy}{dx}=\frac{2\lambda}{y}.dxdy​=y2λ​.

At P=(λ,2λ)P=(\lambda,2\lambda)P=(λ,2λ), mp=2λ2λ=1.m_p=\frac{2\lambda}{2\lambda}=1.mp​=2λ2λ​=1.

Thus the tangent to the parabola at PPP has slope 111.

  1. Find the tangent slope to the ellipse at PPP

Differentiate x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1a2x2​+b2y2​=1 implicitly: 2xa2+2yb2dydx=0.\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0.a22x​+b22y​dxdy​=0.

Hence dydx=−b2xa2y.\frac{dy}{dx}=-\frac{b^2x}{a^2y}.dxdy​=−a2yb2x​.

At P=(λ,2λ)P=(\lambda,2\lambda)P=(λ,2λ), me=−b2λa2(2λ)=−b22a2.m_e=-\frac{b^2\lambda}{a^2(2\lambda)}=-\frac{b^2}{2a^2}.me​=−a2(2λ)b2λ​=−2a2b2​.

  1. Use perpendicularity of tangents

If two tangents are perpendicular, then product of slopes is −1-1−1: mp me=−1.m_p\,m_e=-1.mp​me​=−1.

So,

\implies \frac{b^2}{2a^2}=1 \implies b^2=2a^2.$$ Thus, $$\frac{b^2}{a^2}=2.$$ So the major axis is along $y$-direction. For the ellipse, - semi-major axis $=b$ - semi-minor axis $=a$ Its eccentricity is $$e=\sqrt{1-\frac{a^2}{b^2}}.$$ Using $b^2=2a^2$, $$e=\sqrt{1-\frac{a^2}{2a^2}}= \sqrt{1-\frac12}= \sqrt{\frac12}= rac1{\sqrt2}.$$ --- 6. **Check with options** $$e=\frac{1}{\sqrt2}$$ which is **Option A**.
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