Parabola practice · Mixed difficulty

Option A ($x y=\frac{105}{64}$) is correct because finding the vertex of the parabola and eliminating the parameter 'a' leads to the given locus.

Option A. $C_1$ and $C_2$ touch each other at one point is incorrect. The correct answer is B ($C_1$ and $C_2$ touch each other exactly at two points) because the curves intersect at two points and have the same tangent at each point.

Option A ($h > 2$) is correct because for three distinct normals to be drawn from a point on the axis, the x-coordinate of the point must be greater than 2.

Option B ($x=2 t^2, y=4 t$) is correct because substituting these parametric equations into $y^2=8x$ yields a true statement.

Option C (${(x + 1)}^2 = 4(y - 1)$) is correct because it represents the mirror image of the given parabola after reflection in the tangent line.

Option A ($x+2y=4$) is correct because it represents the directrix, which is perpendicular to the axis and equidistant from the vertex as the focus.

Option A ($x = \pm \frac{{{a^2}}}{{\sqrt {{a^2} - {b^2}} }}$) is correct because the locus of the intersection of tangents at the endpoints of a focal chord is the directrix of the ellipse.

Option D ($y=x+2$) is correct because by finding the tangent to the parabola and then applying the condition for tangency to the hyperbola, we arrive at this equation.

Option B (0) is correct because the sum of the ordinates of the feet of the normals from any point to the parabola ${y^2} = 8x$ is zero.

Option A ($\left\{ { - 1,1} \right\}$) is correct because these are the possible slopes of the focal chord that is tangent to the given circle.

Option A ($x=\frac{{3}}{{4}}$) is correct because the axis of the parabola is found by completing the square and identifying the x-coordinate of the vertex.

Option D (${y^2} - 4x + 2 = 0$) is correct because it represents the locus of the midpoint of PQ, derived by substituting the coordinates of a general point on the parabola into the midpoint formula.

Option C ($x+y+1=0$) is correct because it is the equation of the tangent to the parabola $y^2=4x$ at the point (1, -2).

Option D (${90^0}$) is correct because the normals at the ends of the latus rectum of a parabola are always perpendicular.

Option D ($y=x-1$) is correct because it represents the tangent line to the parabola at x=1, found by calculating the derivative and using the point-slope form.

Option B ((3,0)) is correct because the normals at the ends of the latus rectum of the parabola $y^2=4x$ intersect at the point (3,0).

Option D ((18,-12)) is correct because it represents the coordinates of the point where the normal to the parabola at (2,4) meets the parabola again, calculated using the properties of the parabola and normal.

Option A (48) is correct because it represents the area of the triangle formed by the points of tangency and the focus of the parabola.

Option B ($\pi / 3$) is correct because the angle between the tangents is found by solving a quadratic equation and using the tangent formula.

Option C ($\frac{16}{25}$) is correct because it is the value of k that satisfies the given conditions for the intersection points of the two parabolas and the length of the line segment AB.

Option C (-128) is correct because it is the value of b that satisfies the condition that the line y=mx+4 is a common tangent to both parabolas.

Option C (${\tan ^{ - 1}}\frac{1}{3}$) is correct because it is the angle between the tangents drawn from the given point to the parabola.

Option C (R) is correct because the condition for the tangent to be normal to the circle is satisfied for all real values of 'a'.

Option D (${y^2} = 2x$) is correct because the locus of the midpoint of a chord passing through the vertex of the parabola ${y^2} = 4x$ is ${y^2} = 2x$.

Option D ($y = x + 2$) is correct because it is the common tangent to both the parabola ${y^2} = 8x$ and the hyperbola $xy = -1$, found by equating the tangent equation of the parabola to the hyperbola and solving for the tangency condition.

Option C (x-y-1=0) is correct because it represents the locus of point P, derived from the given condition on the inclinations of the tangents.

Option D (($\frac{{-1}}{{2}}, 0$) $\cup$ ($0, \frac{{1}}{{2}}$)) is correct because it defines the range of 'a' for which three normals can be drawn from the given point to the parabola.

Option A (x + y = 3) is correct because it is the equation of the normal to the parabola that passes through the given point (3,0).

Option B ($3 \sqrt{2} a$) is correct because the length of the common tangent is calculated by finding the points of tangency and then using the distance formula.

Option B A parabola whose vertex is $\left( {3,4} \right)$ and directrix is $x = \frac{{11}}{4}$ is correct because the given equation can be rewritten in the standard form of a parabola, and its vertex and directrix can be determined.