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A family of chords of the parabola \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 \mathrm{\mathrm{c}} ; The locus of their middle point is the curve  k \mathrm{\left(y^2-4 a x\right)(y+2 a)^2-2 a^2 c^2=0} where k =

 

Option: 1

1


Option: 2

2


Option: 3

\sqrt{2}


Option: 4

3


Answers (1)

best_answer

Let the equation of straight line (h,k)  as its mid point. 

then \mathrm{\frac{\mathrm{x}-\mathrm{h}}{\cos \theta}=\frac{\mathrm{y}-\mathrm{k}}{\sin \theta}=\mathrm{r} }.....................(1)
Any point on the line (1) is
\mathrm{(h+r \cos \theta, k+r \sin \theta) }
Solving with the equation of parabola.
\mathrm{\begin{aligned} & y^2=4 a x, \text { we get } \\ &(k+r \sin \theta)^2=4 a(h+r \cos \theta) \\ & \Rightarrow \quad r^2 \sin ^2 \theta+2 r(k \sin \theta-2 a \cos \theta)+k^2-4 a h=0 \end{aligned} }................(2)
which is quadratic in r.

The roots of the quadratic equation will be equal but of opposite sign as ( h, k) is the mid point
\mathrm{\begin{aligned} & \quad \text { Sum of roots }=-\frac{2(k \sin \theta-2 a \cos \theta)}{\sin ^2 \theta}=0 \\ & \therefore \quad k \sin \theta-2 a \cos \theta=0 \\ & \therefore \quad \tan \theta=\frac{2 a}{k} \\ & \text { Now from }(2), \\ & \quad r^2 \sin ^2 \theta+\left(k^2-4 a h\right)=0 \\ & \quad r^2 \cdot \frac{4 a^2}{\left(4 a^2+k^2\right)}+\left(k^2-4 a h\right)=0 \end{aligned} }..............(3)
Length of the chord will be \mathrm{2 \mathrm{r} }. Angle between the two lines will be \mathrm{(\theta-\pi / 4) } and the projection of the chord on the given line will be \mathrm{2 \mathrm{r} \cos (\theta-\pi / 4)=\mathrm{c} }
\mathrm{\begin{array}{ll} \Rightarrow \quad & \frac{2 \mathrm{r}}{\sqrt{2}}(\cos \theta+\sin \theta)=\mathrm{c} \\ \Rightarrow \quad & \frac{2 \mathrm{r}}{\sqrt{2}}\left(\frac{\mathrm{k}+2 \mathrm{a}}{\sqrt{4 \mathrm{a}^2+\mathrm{k}^2}}\right)=\mathrm{c} \\ \Rightarrow \quad & 2 \mathrm{r}^2(\mathrm{k}+2 \mathrm{a})^2=\mathrm{c}^2\left(4 \mathrm{a}^2+\mathrm{k}^2\right) \\ \Rightarrow \quad & \frac{2 \mathrm{r}^2}{\left(4 \mathrm{a}^2+\mathrm{k}^2\right)}=\frac{\mathrm{c}^2}{(k+2 \mathrm{a})^2} \end{array} }
from (3) \& (4) we get
\mathrm{\begin{gathered} \frac{2 a^2 c^2}{(k+2 a)^2}+\left(k^2-4 a h\right)=0 \\ \Rightarrow \quad\left(k^2-4 a h\right)(k+2 a)^2+2 a^2 c^2=0 \end{gathered} }
hence the locus of the middle point is
\mathrm{\left(y^2-4 a x\right)(y+2 a)^2+2 a^2 c^2=0 }
 

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manish

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