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Standard XII

Mathematics

Focal Chord

Let a , r , s...

Question

Let $a, r, s, t$ be nonzero real numbers. Let $P (a t^{2}, 2 a t), Q, R (a r^{2}, 2 a r) and S (a s^{2}, 2 a s)$ be distinct points on the parabola $y^{2} = 4 a x$ . Suppose that $P Q$ is the focal chord and lines $Q R$ and $P K$ are parallel, where $K$ is the point $(2 a, 0) .$

The value of $r$ is

- \frac{1}{t}

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\frac{t^{2} + 1}{t}

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\frac{1}{t}

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\frac{t^{2} - 1}{t}

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Solution

The correct option is D $\frac{t^{2} - 1}{t}$
We know that point $Q$ is,
$= (\frac{a}{t^{2}}, - \frac{2 a}{t})$

Since, $Q R$ is parallel to $P K$ , hence there slope must be same.
$∴ \frac{2 a t - 0}{a t^{2} - 2 a} = \frac{2 a r + \frac{2 a}{t}}{a r^{2} - \frac{a}{t^{2}}} \Rightarrow \frac{t}{t^{2} - 2} = \frac{r + \frac{1}{t}}{r^{2} - \frac{1}{t^{2}}} \Rightarrow \frac{t^{2} - 2}{t} = \frac{r^{2} - \frac{1}{t^{2}}}{r + \frac{1}{t}} \Rightarrow \frac{t^{2} - 2}{t} = r - \frac{1}{t} ∴ r = \frac{t^{2} - 1}{t}$

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Q. Let

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P (a t^{2}, 2 a t), Q, R (a r^{2}, 2 a r)

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be distinct points on the parabola

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Let $a, r, s$ and $t$ be non-zero real numbers. Let $P (a t^{2}, 2 a t), Q, R (a r^{2}, 2 a r)$ and $S (a s^{2}, 2 a s)$ be distinct points on the parabola $y^{2} = 4 a x$ . Suppose that $P Q$ is the focal chord and lines $Q R$ and $P K$ are parallel, where $K$ is point $(2 a, 0)$ .
If $s t = 1$ , then the tangent at $P$ and the normal at $S$ to the parabola meet at a point whose ordinate is

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Focal Chord

Standard XII Mathematics

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Let a,r,s,t be nonzero real numbers. Let P(at2,2at),Q,R(ar2,2ar) and S(as2,2as) be distinct points on the parabola y2=4ax. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point (2a,0). The value of r is

The correct option is D t2−1tWe know that point Q is, =(at2,−2at) Since, QR is parallel to PK, hence there slope must be same. ∴2at−0at2−2a=2ar+2atar2−at2⇒tt2−2=r+1tr2−1t2⇒t2−2t=r2−1t2r+1t⇒t2−2t=r−1t∴r=t2−1t