Mathematics · 2022
JEE Main · 25 June 2022, Shift 2 · Q29
Let the eccentricity of the hyperbola x^2/a^2-y^2/b^2=1 be 5/4. If the equation of the normal at the point (8/(√ 5), 12/5) on the hyperbola is 8 √ 5…
Let the eccentricity of the hyperbola $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be $\displaystyle \frac{5}{4}$. If the equation of the normal at the point $\displaystyle \left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$ on the hyperbola is $\displaystyle 8 \sqrt{5} x+\beta y=\lambda$, then $\displaystyle \lambda-\beta$ is equal to \_\_\_\_.
Official answer
85
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.