Question

If two distinct chords of a parabola x2=4ay passing through (2a,a) are bisected on the line x+y=1, then length of latus rectum can be

Answer 1

Solution:

The given parabola is ,

$x^2= 4 a y$

The given parabola has two distinct chords which passes through (2 a, a) and it is bisected on the line x+y=1.

Since the chord passes through (2 a, a), it will also pass through (-2 a, a).

The two chords is bisected by line , x+ y=1

So ,Mid point of one chord joining (2 a,a) and (-2 a,a) = $[\frac{x_{1} +x_{2}}{2},\frac{y_{1}+y_{2}}{2}]=(\frac{2 a -2a}{2}.\frac{{a+a}}{{2}})=(0,a)$

So, (0,a) passes through , x+y=1

0 +a=1

a=1

So, length of Latus rectum = 4 a= 4 ×1= 4 Units

Answer 2

Answer:

length of latus rectum could take any of the values between 0 and 4 .

Step-by-step explanation:

Hi,

Given that the chords of the parabola  x² = 4ay are bisected by the line

x + y = 1,

Any point on the given line x + y = 1 will be of the form (k, 1 - k).

So, let the midpoint of the chord be M ( k, 1 - k)

The equation of the chord to parabola S when midpoint is known is given by T = S₁

Hence, the equation of the chord whose midpoint is (k, 1 - k) wil be

kx - 2a(y + 1 - k) = k² - 4a(1 - k)

But this chord passes through (2a, a)

⇒2ak - 2a(a + 1 - k) = k² - 4a(1 - k)

⇒k² + 2a² - 2a = 0

⇒k² = 2a - 2a²

The above equation in k should have 2 distinct roots, since 2 chords are possible.

Hence 2a - 2a² > 0

⇒2a(1 - a) > 0

⇒ a ∈ (0, 1)

Thus, 4a ∈ (0, 4)

Hence, length of latus rectum could take any of the values between 0 and 4 .

Hope, it helps !