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NCERT Exemplar · Q38

Q.If the parabola y2=4axy^2 = 4ax passes through the point (3,2)(3, 2), then the length of its latus rectum is
(A) 23\dfrac{2}{3}
(B) 43\dfrac{4}{3}
(C) 13\dfrac{1}{3}
(D) 44

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Substitute the given point into the parabola equation to find aa, then use the fact that the latus rectum of y2=4axy^2 = 4ax has length 4a4a. The answer is 43\boxed{\frac{4}{3}}.

The parabola y2=4axy^2 = 4ax is in standard form with vertex at the origin and axis along the positive xx-axis. The parameter aa controls how "wide" or "narrow" the parabola opens. The latus rectum is the chord through the focus perpendicular to the axis, and for this standard form its length is always 4a4a. So our task reduces to finding aa.

When we say the parabola "passes through" a point, we mean that point satisfies the equation. Substituting (3,2)(3, 2) into y2=4axy^2 = 4ax will give us the value of aa.

Step-by-step solution:

  1. Substitute the point (3,2)(3, 2) into the equation.

    We have x=3x = 3 and y=2y = 2. Plugging these into y2=4axy^2 = 4ax:

22=4a⋅32^2 = 4a \cdot 3

4=12a4 = 12a

  1. Solve for aa.

    Dividing both sides by 1212:

a=412=13a = \frac{4}{12} = \frac{1}{3}

  1. Find the length of the latus rectum.

    For the parabola y2=4axy^2 = 4ax, the length of the latus rectum is 4a4a. Substituting a=13a = \frac{1}{3}: …

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