Let $\mathrm{P}(3,3)$ be a point on the hyperbola, $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$. If the normal to it at P intersects the $x$-axis at $(9,0)$ and $e$ is its eccentricity, then the ordered pair $\left(a^{2}, e^{2}\right)$ is equal to:

(D) $\left(\frac{9}{2}, 3\right)$

(A) $\left(\frac{3}{2}, 2\right)$

(B) $\left(\frac{9}{2}, 2\right)$

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Hyperbola

Let

P (3, 3)

be a point on the hyperbola,

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

. If the normal to it at P intersects the

x

-axis at

(9, 0)

and

e

is its eccentricity, then the ordered pair

(a^{2}, e^{2})

is equal to:

Answer: (D)

Since

P (3, 3)

lies on the hyperbola,

\frac{9}{a ^{2}} - \frac{9}{b ^{2}} = 1

. The tangent slope at

P

\frac{b ^{2} x}{a ^{2} y} = \frac{b ^{2}}{a ^{2}}

, so the normal has slope

- \frac{a ^{2}}{b ^{2}}

and its equation is

y - 3 = - \frac{a ^{2}}{b ^{2}} (x - 3)

. Passing through

(9, 0)

gives

- 3 = - \frac{a ^{2}}{b ^{2}} (6)

, so

\frac{a ^{2}}{b ^{2}} = \frac{1}{2}

, i.e.

b^{2} = 2 a^{2}

. Substituting into the point condition,

\frac{9}{a ^{2}} - \frac{9}{2 a ^{2}} = 1

, so

\frac{9}{2 a ^{2}} = 1

and

a^{2} = \frac{9}{2}

. The eccentricity follows from

e^{2} = 1 + \frac{b ^{2}}{a ^{2}} = 1 + 2 = 3

. Thus

(a^{2}, e^{2}) = (\frac{9}{2}, 3)

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option