Some problems on conics (parabola, ellipse, hyperbola) : IITJEE Mains

Some problems on conics (parabola, ellipse, hyperbola) : IITJEE Mains — Basics 2

I am solving some “nice” problems below:

Problem 1:

Show that the locus of a point that divides a chord of slope 2 of the parabola $y^{2}=4x$ internally in the ratio $1:2$ is a parabola. Find the vertex of this parabola.

Solution 1:

Let $P(t_{1}^{2}, 2t_{1})$ and $Q(t_{2}^{2}, 2t_{2})$ be the extremities of the chord with slope 2.

Hence, $\frac{2t_{1}-2t_{2}}{t_{1}^{2}-t_{2}^{2}}=2 \Longrightarrow t_{1}+t_{2}=1$

Let $(t_{1},t_{2})$ be co-ordinates of the point which divides PQ in the ratio $1:2$ . Then,

$h=\frac{2t_{1}^{2}+t_{2}^{2}}{3}$ and $k=\frac{4t_{1}+2t_{2}}{3}$

$\Longrightarrow 3h = 2t_{1}^{2}+(1-t_{1})^{2}$ and $3k = 4t_{1}+2(1-t_{1})$

$\Longrightarrow 3h = 3t_{1}^{2} - 2t_{1} + 1$ and $3k = 2t_{1}+2$

$\Longrightarrow 3h = 3(\frac{3k-2}{2})^{2} -2(\frac{4k-2}{2})+1$

$\Longrightarrow 12h = 3(9k^{2}-12k+4)-12k+8+4$

$4h = 9k^{2}-16k+8$

Hence, the locus of $(h,k)$ is $9y^{2}-16y-4x+8=0$

$\Longrightarrow (3y-\frac{8}{3})^{2}=4x-8+\frac{64}{9}$

$\Longrightarrow (y-\frac{8}{9})^{2}=\frac{4}{9}(x-\frac{2}{9})$ , which is a parabola with vertex $(\frac{2}{9}, \frac{8}{9})$ .

Problem 2:

If $P_{1}P_{2}$ and $P_{3}P_{4}$ are two focal chords of the parabola $y^{2}=4ax$ , then show that the chords $P_{1}P_{3}$ and $P_{2}P_{4}$ intersect on the directrix of the parabola.

Solution 2:

Let the co-ordinates of $P_{i}$ be $(at_{i}^{2},2at_{i})$ for $i=1,2,3,4$ .

Since $P_{1}P_{2}$ is a focal chord, $t_{1}t_{2}=-1$ —– call this Equation I.

Similarly, $t_{3}t_{4}=-1$ —- call this Equation II.

Equation of $P_{1}P_{3}$ is $y(t_{1}+t_{3})=2(x+at_{1}t_{3})$ —- call this Equation III.

and that of $P_{2}P_{4}$ is $y(t_{2}+t_{4})=2(x+\alpha t_{2}t_{4})$ —- call this Equation IV.

Using I and II, IV reduces to $y(-\frac{1}{t_{1}}-\frac{1}{t_{2}})=2(x+\frac{a}{t_{1}}t_{3})$

that is, $-y(t_{1}+t_{3})=2(xt_{1}t_{3}+a)$ — call this Equation V.

Adding III and V we get:

$0=2[x+at_{1}t_{3}+xt_{1}t_{3}+a]$ , which in turn implies, $(x+a)(1+t_{1}t_{3})=0$ , which in turn implies, that $x=-a$ . Hence, III and V intersect on the directrix $x+a=0$ .

More later,

Nalin Pithwa.

This entry was written by nkpithwa, posted on April 5, 2017 at 5:19 pm, filed under co-ordinate geometry, IITJEE Mains. Bookmark the permalink. Follow any comments here with the RSS feed for this post. Post a comment or leave a trackback: Trackback URL.

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