Parabola practice · Mixed difficulty

Option A ($\frac{{25}}{2}$) is correct because it is the area of the triangle formed by the tangent, normal, and parabola at the specified points.

Option A (3) is correct because the angle between the tangents is found using the locus of the point of intersection of tangents and the given parabola equation.

Option B (9) is correct because substituting the normal form into the given equation yields k = 9.

Option C ($c=-a m^2$) is correct because it is the condition for the line to be tangent to the parabola.

Option A ($\frac{{24}}{{5}}$) is correct because the latusrectum is twice the semi-latusrectum, which is the harmonic mean of the segments of the focal chord.

Option B ($b=\sqrt{2} a$) is correct because it provides the relationship between a and b when the tangent to the parabola and circle is common and makes an angle of $\frac{{\pi}}{{4}}$ with the x-axis.

Option D ($(2,6)$) is correct because substituting y = 3x into the parabola equation yields x = 2 and y = 6.

Option B ($3 x+4 y-1=0$) is correct because the given equation is in the form SP = PM, where PM is the perpendicular distance from the point to the directrix.

Option A $\left(x_1, y_1\right)$ is correct because the midpoint of the chord QR is the same as the point P on the original parabola.

Option A ((1,1)) is correct because it represents the vertex of the parabola defined by the given parametric equations.

Option A (${h^2} = ab$) is correct because this is the condition for the given general equation to represent a parabola.

Option A (18) is correct because substituting the given point into the parabola equation allows us to solve for 'a', and then calculate the latus rectum using 4a.

Option B (its directrix) is correct because the locus of the point of intersection of perpendicular tangents to a parabola is its directrix.

Option A ($\pm \, \frac{\pi }{2}$) is correct because the difference between the eccentric angles of the ends of a pair of conjugate diameters of an ellipse is $\pm \, \frac{\pi }{2}$.

Option D ((-6,-7)) is correct because the tangent to the parabola at (1,7) intersects the circle at (-6,-7).

Option B ($x + a = 0$) is correct because the locus of the intersection of perpendicular tangents to a parabola is its directrix.

Option A (${x^2} + {y^2} - 2xy + 8x + 8y - 16 = 0$) is correct because it represents the equation of a parabola derived from the definition where the distance to the focus equals the distance to the directrix.

The correct answer is ${x^2} - {y^2} - 6ax + 9{a^2} = 0$ because it represents the combined equation of the normals at the ends of the latus rectum of the given parabola.

Option A (8) is correct because the minimum value of $t_1^2$ is found using the relationship between $t$ and $t_1$ and applying AM-GM inequality.

Option A (8) is correct because after finding the incenter and the 'a' value of the parabola, we calculate the focus and then compute the required ratio.

Option C (${a^2}{l^2} + {b^2}{m^2} = 2{n^2}$) is correct because it is derived by comparing the given line with the equation of the chord joining two points on the ellipse with eccentric angles differing by 90 degrees.

Option D (4) is correct because the length of the latus rectum of a parabola in the form $(x-h)^2 = 4p(y-k)$ is $4p$.

Option B (6) is correct because the distance of the intersection point of tangents from the focus is the geometric mean of the distances of P and Q from the focus.

Option C ($y^2-2 a x+2 a^2=0$) is correct because it is the locus of the midpoints of the chords of the parabola that pass through the focus, found by using the equation T = S' and the condition that the chord passes through the focus.

Option A (1) is correct because the abscissa (x-coordinate) of the point on the parabola is 1, derived from the focal distance formula.

Option C ($4 \sqrt{2}$) is correct because the length of the latus rectum is four times the distance from the focus to the tangent at the vertex.

Option D ($32 \sqrt{2}$) is correct because the length of the chord through L with slope -1 is calculated to be $32 \sqrt{2}$.

Option A ($y=0, y=4(x-1)$) is correct because these lines are tangent to both parabolas, as verified by solving for the tangency conditions.

Option A (3) is correct because the tangent of the angle between the two tangents is calculated using the slopes of the tangents and the formula for the angle between two lines.

Option D ($\frac{5 a}{2}$) is correct because it is the value of c that makes the discriminant of the quadratic equation zero, indicating tangency.