In the line alpha x+4 y=sqrt(7), where alpha in R, touches the ellipse 3 x^2+4 y^2=1 at the point P in the first quadrant, then one of the focal distances of P is :

alpha x+4 y-sqrt(7)=0 touches 3 x^2+4 y^2=1 Therefore c^2= a^2 m^2+ b^2 (7 / 16)=(1 / 3) x fracalpha^216+(1 / 4) => alpha=3,-3 Tangent is 3 x+4 y-sqrt(7)=0 Let the point of contact is P( x_1 y_1) ∴ Tangent is 3 xx_1+4 yy_1=1 Therefore frac3 x_13=frac4 y_14=frac1sqrt(7) Therefore P(frac1sqrt(7), frac1sqrt(7)) e=sqrt(1-(3 / 4))=(1 / 2) \ PS= e( PM) =e(frac a e-frac1sqrt(7)) =(1 / 2)(frac2sqrt(3)-frac1sqrt(7))=frac1sqrt(3)-frac12 sqrt(7) PS^prime= e( PM^prime)=(1 / 2)(frac a e+frac1sqrt(7))=(1 / 2)(frac1sqrt(7)+frac2sqrt(3)) =frac1sqrt(3)+frac12 sqrt(7)

In the line alpha x 4 y sqrt 7 where alpha in R touches... | null (Mathematics)

\alpha x+4 y-\sqrt{7}=0

touches

3 x^{2}+4 y^{2}=1\\

\therefore \mathrm{c}^{2}=\mathrm{a}^{2} \mathrm{~m}^{2}+\mathrm{b}^{2}\\

\frac{7}{16}=\frac{1}{3} \times \frac{\alpha^{2}}{16}+\frac{1}{4} \Rightarrow \alpha=3,-3\\

Tangent is

3 x+4 y-\sqrt{7}=0\\

Let the point of contact is

\mathrm{P}\left(\mathrm{x}_{1} \mathrm{y}_{1}\right)\\

∴ Tangent is

3 \mathrm{xx}_{1}+4 \mathrm{yy}_{1}=1\\

\therefore \frac{3 \mathrm{x}_{1}}{3}=\frac{4 \mathrm{y}_{1}}{4}=\frac{1}{\sqrt{7}} \quad \therefore \mathrm{P}\left(\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right)\\

\mathrm{e}=\sqrt{1-\frac{3}{4}}=\frac{1}{2}\\

\\ \mathrm{PS}=\mathrm{e}(\mathrm{PM})\\

=e\left(\frac{\mathrm{a}}{\mathrm{e}}-\frac{1}{\sqrt{7}}\right)\\

=\frac{1}{2}\left(\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{7}}\right)=\frac{1}{\sqrt{3}}-\frac{1}{2 \sqrt{7}}\\

\mathrm{PS}^{\prime}=\mathrm{e}\left(\mathrm{PM}^{\prime}\right)=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{e}}+\frac{1}{\sqrt{7}}\right)=\frac{1}{2}\left(\frac{1}{\sqrt{7}}+\frac{2}{\sqrt{3}}\right)\\

=\frac{1}{\sqrt{3}}+\frac{1}{2 \sqrt{7}}

Explanation

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