Let the tangent drawn to the parabola y^2=24x at the point (,) be perpendicular to the lin…

Question

Let the tangent drawn to the parabola $y^2=24x$ at the point $(\alpha,\beta)$ be perpendicular to the line $2x+2y=5$. Then the normal to the hyperbola $\dfrac{x^2}{\alpha^2}-\dfrac{y^2}{\beta^2}=1$ at the point $(\alpha+4,\beta+4)$ does NOT pass through the point:

Accepted Answer

Tangent at $(\alpha,\beta)$: $\beta y=12(x+\alpha)$, slope $=12/\beta$. Perpendicular to slope $-1$ means tangent slope $=1$: $\beta=12,\ \alpha=\beta^2/24=6$.

Hyperbola: $\dfrac{x^2}{36}-\dfrac{y^2}{144}=1$; normal at $(10,16)$:
$$\frac{36x}{10}+\frac{144y}{16}=180\ \Longrightarrow\ \frac x{50}+\frac y{20}=1.$$

Checking each option, $(15,13)$ gives $0.3+0.65=0.95\ne1$, so it does not lie on the normal.

Solutions · 1