If the image of the parabola y=x2 in the line x+y=1 is x+y2=k y, then k is equal to

Question

If the image of the parabola y=x2 in the line x+y=1 is x+y2=k y, then k is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

General point on y=x2 is (t, t2) . Let us find the image of (t, t2) in 4 y=2 x+3 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/cf95d55212301383be17b0141f6c7677-.png" /> (a+t/2)+(b+t2/2)=1 a+b+b+t2=2 and (b-t2/a-t)=1 Solving we get a+b2=2 b ⇒ x+y2=2 y ⇒ k=2

Q. If the image of the parabola $y = x^{2}$ in the line $x + y = 1$ is $x + y^{2} = k y$ , then $k$ is equal to

General point on $y = x^{2}$ is $(t, t^{2}) .$
Let us find the image of $(t, t^{2})$ in $4 y = 2 x + 3$

$\frac{a + t}{2} + \frac{b + t ^{2}}{2} = 1$
$a + b + b + t^{2} = 2$
and $\frac{b - t ^{2}}{a - t} = 1$
Solving we get $a + b^{2} = 2 b$
$\Rightarrow x + y^{2} = 2 y$
$\Rightarrow k = 2$

Q. If the image of the parabola y=x2 in the line x+y=1 is x+y2=ky, then k is equal to

General point on y=x2 is (t,t2). Let us find the image of (t,t2) in 4y=2x+3 2a+t​+2b+t2​=1 a+b+b+t2=2 and a−tb−t2​=1 Solving we get a+b2=2b ⇒x+y2=2y ⇒k=2

Q. If the image of the parabola $y = x^{2}$ in the line $x + y = 1$ is $x + y^{2} = k y$ , then $k$ is equal to

General point on $y = x^{2}$ is $(t, t^{2}) .$
Let us find the image of $(t, t^{2})$ in $4 y = 2 x + 3$

$\frac{a + t}{2} + \frac{b + t ^{2}}{2} = 1$
$a + b + b + t^{2} = 2$
and $\frac{b - t ^{2}}{a - t} = 1$
Solving we get $a + b^{2} = 2 b$
$\Rightarrow x + y^{2} = 2 y$
$\Rightarrow k = 2$