If the curves (x2/c)+(y2/4)=1 and y3=16 x intersect at right angles, then c is equal to

Question

If the curves (x2/c)+(y2/4)=1 and y3=16 x intersect at right angles, then c is equal to (A) (3/4) (B) (1/2) (C) (4/3) (D) 2

Tardigrade Admin · Accepted Answer

Let (x1, y1) be their point of intersection. So, m 1 × m 2=-1 ⇒((64 x 1/ c ))=3 y 1 3...(1) Also, y 1 3=16 x 1....(2) ∴ From (1) and (2), we get c=(4/3)

Q. If the curves $\frac{x ^{2}}{c} + \frac{y ^{2}}{4} = 1$ and $y^{3} = 16 x$ intersect at right angles, then $c$ is equal to

$\frac{3}{4}$

$\frac{1}{2}$

$\frac{4}{3}$

2

Let $(x_{1}, y_{1})$ be their point of intersection.
So, $m_{1} \times m_{2} = - 1 \Rightarrow (\frac{64 x _{1}}{c}) = 3 y_{1}^{3}$ ...(1)
Also, $y_{1}^{3} = 16 x_{1}$ ....(2)
$∴$ From (1) and (2), we get $c = \frac{4}{3}$

Q. If the curves cx2​+4y2​=1 and y3=16x intersect at right angles, then c is equal to

43​

21​

34​

2

Let (x1​,y1​) be their point of intersection. So, m1​×m2​=−1⇒(c64x1​​)=3y1​3...(1) Also, y1​3=16x1​....(2) ∴ From (1) and (2), we get c=34​

Q. If the curves $\frac{x ^{2}}{c} + \frac{y ^{2}}{4} = 1$ and $y^{3} = 16 x$ intersect at right angles, then $c$ is equal to

$\frac{3}{4}$

$\frac{1}{2}$

$\frac{4}{3}$

Let $(x_{1}, y_{1})$ be their point of intersection.
So, $m_{1} \times m_{2} = - 1 \Rightarrow (\frac{64 x _{1}}{c}) = 3 y_{1}^{3}$ ...(1)
Also, $y_{1}^{3} = 16 x_{1}$ ....(2)
$∴$ From (1) and (2), we get $c = \frac{4}{3}$