Parabola Examples

PARABOLA EXAMPLES

Example 1 : The length of latus rectum of a parabola, whose focus is (2, 3) and directrix is the line x - 4y + 3 = 0 is -

Solution : The length of latus rectum = 2 x perp. from focus to the dirctrix

= 2 x |\({2-4(3)+3}\over {\sqrt{1+16}}\)| = \(14\over \sqrt{17}\)



Example 2 : Find the value of k for which the point (k-1, k) lies inside the parabola \(y^2\) = 4x.

Solution : \(\because\)     Point (k-1, k) lies inside the parabola \(y^2\) = 4x.

\(\therefore\)     \({y_1}^2 - 4ax_1\) < 0

\(\implies\)     \(k^2\) - 4(k-1) < 0

\(\implies\)     \(k^2\) - 4k + 4 < 0

\((k-2)^2\) < 0 \(\implies\) k \(\in\) \(\phi\)



Example 3 : Find the equation of the tangents to the parabola \(y^2\) = 9x which go through the point (4,10).

Solution : Equation of tangent to the parabola \(y^2\) = 9x is

y = mx + \(9\over 4m\)

Since it passes through (4,10)

\(\therefore\)     10 = 4m + \(9\over 4m\) \(\implies\) 16\(m^2\) - 40m + 9 = 0

m = \(1\over 4\), \(9\over 4\)

\(\therefore\)     Equation of tangent's are y = \(x\over 4\) + 9 & y = \(9x\over 4\) + 1



Example 4 : Find the locus of middle point of the chord of the parabola \(y^2\) = 4ax which pass through a given (p,q).

Solution : Let P(h,k) be the mid point of chord of the parabola \(y^2\) = 4ax,

so equation of chord is yk - 2a(x+h) = \(k^2\) - 4ah.

Since it passes through (p,q)

\(\therefore\)     qk - 2a(p+h) = \(k^2\) - 4ah

\(\therefore\)     Required locus is \(y^2\) - 2ax - qy + 2ap = 0


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