Parabola Questions

Solution : The given parabola is of form $y^2$ = 4ax. On comparing, we have 4a = 12 i.e a = 3. We know that the focal distance of any point (x, y) on $y^2$ = 4ax is x + a. Let the given point on the parabola $y^2$ = 12 x be (x, y). […]

Solution : The equation of tangent to the parabola $y^2$ = 4ax is y = mx + $a\over m$. Since it is drawn from point (2,3) Therefore it lies on tangent y = mx + $a\over m$. $\implies$ 3 = 2m + $a\over m$ $\implies$ 3m = 2$m^2$ + a $\implies$  2$m^2$ – 3m +

Solution : for parabola, $y^2$ = 4x Let y = mx + $1\over m$ is tangent line and it touches the parabola $x^2$ = -32. $\therefore$ $x^2$ = -32(mx + $1\over m$) $\implies$ $x^2 + 32mx + {32\over m}$ = 0 Now, D = 0 because it touches the curve. $\therefore$ $(32m)^2 – 4.{32\over m}$

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 Similar

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

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 $\in$ $\phi$ Similar Questions The slope of the line touching both the parabolas $y^2$ = 4x and

Solution : The length of latus rectum = 2 x perpendicular from focus to the directrix = 2 x |${2-4(3)+3}\over {\sqrt{1+16}}$| = $14\over \sqrt{17}$ Similar Questions The slope of the line touching both the parabolas $y^2$ = 4x and $x^2$ = -32 is Find the locus of middle point of the chord of the parabola

Solution : The equation of tangent in slope form to $y^2$ = 4ax is y = mx + $a\over m$ Now, if it is common to both parabola, it also lies on second parabola then $x^2$ = 4a(mx + $a\over m$) $mx^2 – 4am^2 – 4a^2$ = 0 has equal roots. then its discriminant is

Solution : The Equation of tangent to the parabola having slope ‘m’, is y = mx + $a\over m$ ,  (m $\ne$ 0) and point of contact  is ($a\over m^2$, $2a\over m$). Similar Questions The sum of the slopes of the tangent of the parabola $y^2$=4ax drawn from the point (2,3) is Find the locus