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A chord of the ellipse \mathrm{16 x^2+9 y^2-32 x-36 y=0} which subtends a right angle at the origin always passes through a fixed point. Find coordinates of that point.

Option: 1

\left(\frac{32}{25}, \frac{36}{25}\right)


Option: 2

\left(\frac{7}{25}, \frac{8}{25}\right)


Option: 3

\left(\frac{17}{25}, \frac{18}{23}\right)


Option: 4

\left(\frac{12}{25}, \frac{13}{25}\right)


Answers (1)

best_answer

Equation of the ellipse is

\mathrm{ 16 x^2+9 y^2-32 x-36 y=0 }
Let equation of the chord be \mathrm{a x+b y=1}

By homogenizing the equation of the ellipse w.r.t equation of the chord we will get the equation of two lines joining the origin and point of intersection of the chord and the ellipse

\mathrm{ \begin{aligned} & 16 x^2+9 y^2-32 x(a x+b y)-36 y(a x+b y)=0 \\\\ & \text { i.e. } x^2(16-32 a)+y^2(9-36 b)-(32 b+36 a) x y=0 \end{aligned} }

Since the two lines are perpendicular to each other

Coeff of  \mathrm{x^2+} coeff. of \mathrm{y^2=0}

\mathrm{ 16-32 a+9-36 b=0 }

\mathrm{ 32 a+36 b=25 \text { i.e. } \frac{32}{25} a+\frac{36}{25} b=1 }
Comparing the above condition with the equation of the chord \mathrm{a x+b y=1}

It is clear that the chord passes through a fixed point whose coordinates are

\mathrm{ \left(\frac{32}{25}, \frac{36}{25}\right) }

Posted by

Rishi

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