Let P(4,3) be a point on the hyperbola frac{x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.Since $P(4,3)$ lies on the hyperbola, we have $\frac{16}{a^{2}}-\frac

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.Since $P(4,3)$ lies on the hyperbola, we have $\frac{16}{a^{2}}-\frac{9}{b^{2}}=1$  $(i)$.The equation of the normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at $(x_1, y_1)$ is $\frac{a^{2}x}{x_1} + \frac{b^{2}y}{y_1} = a^{2} + b^{2}$.Substituting $(x_1, y_1) = (4,3)$, the normal equation is $\frac{a^{2}x}{4} + \frac{b^{2}y}{3} = a^{2} + b^{2}$.Since the normal passes through $(16,0)$, we substitute $x=16$ and $y=0$:$\frac{a^{2}(16)}{4} + 0 = a^{2} + b^{2}$ $\Rightarrow 4a^{2} = a^{2} + b^{2}$ $\Rightarrow 3a^{2} = b^{2}$.Substitute $b^{2} = 3a^{2}$ into equation $(i)$:$\frac{16}{a^{2}} - \frac{9}{3a^{2}} = 1$ $\Rightarrow \frac{16}{a^{2}} - \frac{3}{a^{2}} = 1$ $\Rightarrow \frac{13}{a^{2}} = 1$ $\Rightarrow a^{2} = 13$.Then $b^{2} = 3(13) = 39$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{39}{13}} = \sqrt{1 + 3} = \sqrt{4} = 2$.

Let $P(4,3)$ be a point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. If the normal at $P$ intersects the $X$-axis at $(16,0)$, then the eccentricity of the hyperbola is

Similar Questions

The angle between the asymptotes of the hyperbola $x^2-3y^2=3$ is

If two tangents are drawn to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that the product of their gradients is $c^2$,then they intersect on the curve:

If the angle between the asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{3} = 4$ is $\frac{\pi}{3}$,then its conjugate hyperbola is:

If a circle of radius $4 \text{ cm}$ passes through the foci of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{4} = 1$ and is concentric with the hyperbola,then the eccentricity of the conjugate hyperbola of that hyperbola is

The locus of a variable point whose chord of contact with respect to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ subtends a right angle at the origin is

Vedclass Test Series

Exam Paper Generator

Online Exam Module