Conic Sections
Conic Section
Allen Star Batch
Grade 11
Question:
If the tangent drawn at point $(t^2, 2t)$ on the parabola $y^2 = 4x$ is same as the normal drawn at point $(\sqrt{5}\cos\theta, 2\sin\theta)$ on the ellipse $4x^2 + 5y^2 = 20$. Then:
$\theta = \cos^{-1}\left(-\frac{1}{\sqrt{5}}\right)$
$\theta = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right)$
$t = -\frac{2}{\sqrt{5}}$
$t = -\frac{1}{\sqrt{5}}$
Step-by-Step Solution
Key Concept: Matching the tangent to a parabola with the normal to an ellipse requires equating coefficients from two linear equations in different parametrizations.
The tangent to the parabola $y^2 = 4x$ at $(t^2, 2t)$ is $ty = x + t^2$. The normal to the ellipse $4x^2 + 5y^2 = 20$ at $(\sqrt{5}\cos\theta, 2\sin\theta)$ is $(\sqrt{5}\sec\theta)x - (2\csc\theta)y = 1$. Setting these lines equal and solving yields $t = -\frac{2}{\sqrt{5}}\cot\theta$ and $t = -\frac{1}{2}\sin\theta$. This leads to $4\cos\theta = -\sqrt{5}\sin^2\theta$, which simplifies to $\sqrt{5}\cos^2\theta - 4\cos\theta - \sqrt{5} = 0$.
Correct Answer: 1,4