If p and q are the x and y-intercepts respect | vedclass

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(C) The equation of the line in intercept form is $\frac{x}{p} + \frac{y}{q} = 1$. Since the line passes through $(a \cos \alpha, b \sin \alpha)$ and

Vedclass · Accepted Answer

$\sec ^2\left(\frac{\alpha-\beta}{2}\right)$

Vedclass · Answer

(C) The equation of the line in intercept form is $\frac{x}{p} + \frac{y}{q} = 1$. Since the line passes through $(a \cos \alpha, b \sin \alpha)$ and $(a \cos \beta, b \sin \beta)$,we have: $\frac{a \cos \alpha}{p} + \frac{b \sin \alpha}{q} = 1$ $(i)$ $\frac{a \cos \beta}{p} + \frac{b \sin \beta}{q} = 1$ (ii) Subtracting (ii) from $(i)$: $\frac{a}{p}(\cos \alpha - \cos \beta) + \frac{b}{q}(\sin \alpha - \sin \beta) = 0$ $\frac{a}{p}(-2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}) = -\frac{b}{q}(2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2})$ $\frac{a}{p} \sin \frac{\alpha+\beta}{2} = \frac{b}{q} \cos \frac{\alpha+\beta}{2} = k$ (let) Then $\frac{a}{p} = k \csc \frac{\alpha+\beta}{2}$ and $\frac{b}{q} = k \sec \frac{\alpha+\beta}{2}$. Substituting into $(i)$: $k \csc \frac{\alpha+\beta}{2} \cos \alpha + k \sec \frac{\alpha+\beta}{2} \sin \alpha = 1$. Solving for $k$ and simplifying the expression $\frac{a^2}{p^2} + \frac{b^2}{q^2}$ leads to $\sec^2 \left(\frac{\alpha-\beta}{2}\right)$.

If $p$ and $q$ are the $x$ and $y$-intercepts respectively of the line passing through the points $(a \cos \alpha, b \sin \alpha)$ and $(a \cos \beta, b \sin \beta)$,then $\frac{a^2}{p^2}+\frac{b^2}{q^2}=$

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