Let $\mathrm{P}(\mathrm{p} \sec \theta, \mathrm{q} \tan \theta)$ and $\mathrm{Q}(\mathrm{p} \sec \phi, \mathrm{q} \tan \phi)$ where $\theta+\phi=\frac{\pi}{2}$ , be two points on the hyperbola $\frac{x^2}{p^2}-\frac{y^2}{q^2}=1$ . If $\left(x_1, y_1\right)$ is the point of intersection of normals at $P$ and $Q$ , then $\mathrm{y}_1$ is equal to

Ellipse

Grade 11

Question:

Let $\mathrm{P}(\mathrm{p} \sec \theta, \mathrm{q} \tan \theta)$ and $\mathrm{Q}(\mathrm{p} \sec \phi, \mathrm{q} \tan \phi)$ where $\theta+\phi=\frac{\pi}{2}$, be two points on the hyperbola $\frac{x^2}{p^2}-\frac{y^2}{q^2}=1$. If $\left(x_1, y_1\right)$ is the point of intersection of normals at $P$ and $Q$, then $\mathrm{y}_1$ is equal to

$\frac{p^2+q^2}{p}$

$\frac{p^2+q^2}{q}$

$-\frac{p^2+q^2}{q}$

$-\frac{p^2+q^2}{p}$

Step-by-Step Solution

Key Concept: Apply the parametric form of the normal to a hyperbola, px cos θ + qy cot θ = p² + q², and solve the system of equations using the complementary angle relationship φ = π/2 - θ.

$-\frac{p^2+q^2}{q}$

Correct Answer: C

Question:

Step-by-Step Solution

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