If 3 x+4 y=12 intersect the ellipse (x2/25)+(y2/16)=1 at P and Q, then the point of intersection of tangents at P and Q is

Question

If 3 x+4 y=12 intersect the ellipse (x2/25)+(y2/16)=1 at P and Q, then the point of intersection of tangents at P and Q is (A) (0,1) (B) (1,-2) (C) ((25/4), (16/3)) (D) (-(25/4), (16/3))

Tardigrade Admin · Accepted Answer

Let point of intersection is R(h, k) P Q is chord of contact ( hx /25)+( ky /16)=1 .....(i) equation (x/4)+(y/3)=1......(ii) Comparing (i) and (ii) ⇒ equation ( h /25)=(1/4), k =(16/3)

Q. If $3 x + 4 y = 12$ intersect the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ at $P$ and $Q$ , then the point of intersection of tangents at $P$ and $Q$ is

$(0, 1)$

$(1, - 2)$

$(\frac{25}{4}, \frac{16}{3})$

$(- \frac{25}{4}, \frac{16}{3})$

Let point of intersection is $R (h, k)$
$PQ$ is chord of contact
$\frac{h x}{25} + \frac{k y}{16} = 1.....$ (i)
equation $\frac{x}{4} + \frac{y}{3} = 1......$ (ii)
Comparing (i) and (ii)
$\Rightarrow$ equation $\frac{h}{25} = \frac{1}{4}, k = \frac{16}{3}$

Q. If 3x+4y=12 intersect the ellipse 25x2​+16y2​=1 at P and Q, then the point of intersection of tangents at P and Q is

(0,1)

(1,−2)

(425​,316​)

(−425​,316​)