The X and Y intercepts of the tangent to the | vedclass

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(B) The equation of the hyperbola is $\frac{x^2}{20}-\frac{y^2}{5}=1$. Here $a^2=20$ and $b^2=5$. The slope of the line $4x+3y=7$ is $m_1 = -\frac{4}{

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$\frac{10}{3}, \frac{-5}{2}$

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(B) The equation of the hyperbola is $\frac{x^2}{20}-\frac{y^2}{5}=1$. Here $a^2=20$ and $b^2=5$. The slope of the line $4x+3y=7$ is $m_1 = -\frac{4}{3}$. Since the tangent is perpendicular to this line,its slope $m = \frac{3}{4}$. The equation of the tangent with slope $m$ is $y = mx \pm \sqrt{a^2m^2-b^2}$. Substituting the values: $y = \frac{3}{4}x \pm \sqrt{20(\frac{9}{16})-5} = \frac{3}{4}x \pm \sqrt{\frac{45}{4}-5} = \frac{3}{4}x \pm \sqrt{\frac{25}{4}} = \frac{3}{4}x \pm \frac{5}{2}$. Case $1$: $y = \frac{3}{4}x + \frac{5}{2} \implies \frac{3}{4}x - y = -\frac{5}{2} \implies \frac{x}{-10/3} + \frac{y}{5/2} = 1$. Intercepts are $-\frac{10}{3}, \frac{5}{2}$. Case $2$: $y = \frac{3}{4}x - \frac{5}{2} \implies \frac{3}{4}x - y = \frac{5}{2} \implies \frac{x}{10/3} + \frac{y}{-5/2} = 1$. Intercepts are $\frac{10}{3}, -\frac{5}{2}$. Comparing with options,option $B$ matches the second case.

The $X$ and $Y$ intercepts of the tangent to the hyperbola $\frac{x^2}{20}-\frac{y^2}{5}=1$ which is perpendicular to the line $4x+3y=7$,are respectively

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