Let L1 be the length of the common chord of the curves x2+y2 = 9 and y2 = 8x, and L2 be the length of the latus rectum of y2 = 8x, then :

Question

Let L1 be the length of the common chord of the curves x2+y2 = 9 and y2 = 8x, and L2 be the length of the latus rectum of y2 = 8x, then : (A) L1 > L2 (B) L1 = L2 (C) L1 < L2 (D) (L1/L2) = √2

Tardigrade Admin · Accepted Answer

x2 + y2 = 9 y2 = 8x L2 = L.R. of y2 = 8x ⇒ L2 = 8 Solve x2 + 8x = 9 ⇒ x = 1,-9 x = -9 reject ∴ y2 = 8 x so y2 = 8 y = ± √8 Point of intersection are (1, √8) (1, - √8) So L1 = 2√8 (L1/L2) = (2√8/8) = (2/√8) = (1/√2) < 1 L1 < L2

Q. Let $L_{1}$ be the length of the common chord of the curves $x^{2} + y^{2} = 9$ and $y^{2} = 8 x,$ and $L_{2}$ be the length of the latus rectum of $y^{2} = 8 x,$ then :

$L_{1} > L_{2}$

$L_{1} = L_{2}$

$L_{1} < L_{2}$

$\frac{L _{1}}{L _{2}} = 2$

Q. Let L1​ be the length of the common chord of the curves x2+y2=9 and y2=8x, and L2​ be the length of the latus rectum of y2=8x, then :

L1​>L2​

L1​=L2​

L1​<L2​

L2​L1​​=2​

x2+y2=9&y2=8x L2​= L.R. of y2=8x⇒L2​=8 Solve x2+8x=9⇒x=1,−9 x=−9reject ∴y2=8x so y2=8 y=±8​ Point of intersection are (1,8​)(1,−8​) So L1​=28​ L2​L1​​=828​​=8​2​=2​1​<1 L1​<L2​

Q. Let $L_{1}$ be the length of the common chord of the curves $x^{2} + y^{2} = 9$ and $y^{2} = 8 x,$ and $L_{2}$ be the length of the latus rectum of $y^{2} = 8 x,$ then :

$L_{1} > L_{2}$

$L_{1} = L_{2}$

$L_{1} < L_{2}$

$\frac{L _{1}}{L _{2}} = 2$