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  • A family of chords of the parabola <img alt="\mathrm{\mathrm{y}^2=4 \mathrm{ax}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cmathrm%7By%7D%5E2%3D4%20%5Cmathrm%7Bax%7D%7D" /

A family of chords of the parabola \mathrm{\mathrm{y}^2=4 \mathrm{ax}} is drawn so that their projections on a straight line inclined equally to both the axes are all of a constant length c; then the locus of their middle point is the curve \mathrm{\left(y^2-4 a x\right)(y+2 a)^2+k a^2 c^2=0}, where \mathrm{ k=}

Option: 1

1


Option: 2

-1


Option: 3

\frac{1}{2}


Option: 4

2


Answers (1)

best_answer

Let the equation of a straight line with \mathrm{(h, k)} as its mid-point be \mathrm{\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=r}

Any point on the above line is \mathrm{\left(x_1+r \cos \theta, y_1+r \sin \theta\right)}

Solving with the equation of the parabola \mathrm{y^2=4 a x,} we get

\mathrm{ r^2 \sin ^2 \theta+2 r(k \sin \theta-2 a \cos \theta)+\left(k^2-4 a h\right)=0 }
which is quadratic in r

The roots of this quadratic equation will be equal but of opposite in sign as \mathrm{(h, k)} is the mid point

\mathrm{\therefore \quad} Coefficient of r is zero

\mathrm{ \Rightarrow \quad k \sin \theta-2 a \cos \theta=0 \Rightarrow \quad \cot \theta=\frac{k}{2 a} }
Now, as coeff of r is zero, we have
\mathrm{ r^2 \sin ^2 \theta=4 a h-k^2 \quad \text { or } \quad r=\frac{\sqrt{4 a h-k^2}}{\sin \theta} }
\mathrm{\therefore \quad}  Length of the chord will be \mathrm{2 r}. Angle between the two lines will be \mathrm{\left(\theta-\frac{\pi}{4}\right)} and the projection of the chord on the given line will be \mathrm{2 r \cos \left(\theta-\frac{\pi}{4}\right)=c.}

\mathrm{ \begin{aligned} & \frac{2 \sqrt{4 a h-\mathrm{k}^2}}{\sin \theta} \cdot \cos \left(\theta-\frac{\pi}{4}\right)=\mathrm{c} \Rightarrow \frac{2 \sqrt{4 \mathrm{ah}-\mathrm{k}^2}}{\sqrt{2}}\left(\frac{\mathrm{k}}{2 \mathrm{a}}+1\right)=\mathrm{c} \\\\ & \left(4 a h-k^2\right)(k+2 a)^2=(\sqrt{2})^2 a^2 c^2 \quad \text { i.e. }\left(k^2-4 a h\right)(k+2 a)^2+2 a^2 c^2=0 \\\\ & \end{aligned} }

Generalising for \mathrm{(\mathrm{h}, \mathrm{k})} we have the required locus as

\mathrm{ \left(y^2-4 a x\right)(y+2 a)^2+a^2 c^2=0 \text {. } }

Posted by

Ritika Harsh

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