Let AB is a normal chord of parabola y 2-4 x- with foot of normal as A 1-2- If AB - p - q - where p is a natural number and q is a prime number- then P - q is equal to

Y^2=4x⇒ a=1 nbsp; Parameter of A:2×1× t_1= 2 ⇒ t_1= 1 nbsp; nbsp;Parameter of B:t_2= t_1 2t_1=3 ∴ B(at_2^2,2at_2)=B(9,6) AB=8√(2) ⇒ p=8, q=2 ⇒ pq=4 ...

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Standard XII

Mathematics

Equation of Normal at Given Point

Let AB is a n...

Question

Let $A B$ is a normal chord of parabola $y^{2} = 4 x$ , with foot of normal as $A (1, - 2)$ . If $A B = p \sqrt{q}$ , where $p$ is a natural number and $q$ is a prime number, then $\frac{p}{q}$ is equal to

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Solution

$y^{2} = 4 x \Rightarrow a = 1$
Parameter of $A : 2 \times 1 \times t_{1} = - 2$
$\Rightarrow t_{1} = - 1$
Parameter of $B : t_{2} = - t_{1} - \frac{2}{t_{1}} = 3$
$∴ B (a t_{2}^{2}, 2 a t_{2}) = B (9, 6)$
$A B = 8 \sqrt{2}$
$\Rightarrow p = 8, q = 2$
$\Rightarrow \frac{p}{q} = 4$

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Let AB is a normal chord of parabola y2=4x, with foot of normal as A(1,−2). If AB=p√q, where p is a natural number and q is a prime number, then pq is equal to

y2=4x⇒a=1 Parameter of A:2×1×t1=−2 ⇒t1=−1 Parameter of B:t2=−t1−2t1=3 ∴B(at22,2at2)=B(9,6) AB=8√2 ⇒p=8,q=2 ⇒pq=4

Let $A B$ is a normal chord of parabola $y^{2} = 4 x$ , with foot of normal as $A (1, - 2)$ . If $A B = p \sqrt{q}$ , where $p$ is a natural number and $q$ is a prime number, then $\frac{p}{q}$ is equal to

$y^{2} = 4 x \Rightarrow a = 1$
Parameter of $A : 2 \times 1 \times t_{1} = - 2$
$\Rightarrow t_{1} = - 1$
Parameter of $B : t_{2} = - t_{1} - \frac{2}{t_{1}} = 3$
$∴ B (a t_{2}^{2}, 2 a t_{2}) = B (9, 6)$
$A B = 8 \sqrt{2}$
$\Rightarrow p = 8, q = 2$
$\Rightarrow \frac{p}{q} = 4$