Hyperbola Ques 27
- If the eccentricity of the standard hyperbola passing through the point $(4,6)$ is 2 , then the equation of the tangent to the hyperbola at $(4,6)$ is (2019 Main, 8 April II)
(a) $3 x-2 y=0$
(b) $x-2 y+8=0$
(c) $2 x-y-2=0$
(d) $2 x-3 y+10=0$
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Answer:
Correct Answer: 27.(c)
Solution:
Formula:
- Let the equation of standard hyperbola is
$$ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $$
Now, eccentricity of hyperbola is
$$ \sqrt{1+\frac{b^{2}}{a^{2}}}=2 $$
$$ \Rightarrow \quad a^{2}+b^{2}=4 a^{2} $$
$\Rightarrow \quad b^{2}=3 a^{2}$
Since, hyperbola (i) passes through the point $(4,6)$
$$ \therefore \quad \frac{16}{a^{2}}-\frac{36}{b^{2}}=1 $$
On solving Eqs. (ii) and (iii), we get
$$ a^{2}=4 \text { and } b^{2}=12 $$
Now, equation of tangent to hyperbola (i) at point $(4,6)$, is
$$ \begin{array}{rlrl} & & \frac{4 x}{a^{2}}-\frac{6 y}{b^{2}}=1 & \\ \Rightarrow \quad & \frac{4 x}{4}-\frac{6 y}{12}=1 \\ \Rightarrow \quad & x-\frac{y}{2}=1 \Rightarrow 2 x-y-2=0 \end{array} \quad \text { [from Eq. (iv)] } $$
BETA
