Question

Match the column 30. Consider the ellipse x² + 2y² = 2; with ends of major axis as A and A'. Point P lying on the ellipse is joined to A & A'. From A' perpendicular is drawn to AP & from A perpendicular is drawn to A'P. Let the locus of their point of intersection be another conic E.

• A. A) Abscissa of points of intersection of given ellipse and conic E is
• B. B) Square root of length of latus rectum of conic E is
• C. C) The abscissa of foci of E is
• D. D) If equation of directrix of E is 2x + y = k, then k is
• E. P) √2
• F. Q) -√2
• G. R) 0
• H. S) 2√2
• I. T) -2√2

Answer 1

A-P, A-Q, B-S, C-R, D-S, D-T

Answer 2

The equation of the given ellipse is $\frac{x^2}{2} + \frac{y^2}{1} = 1$, so $a^2=2, b^2=1$. The ends of the major axis are $A(\sqrt{2}, 0)$ and $A'(-\sqrt{2}, 0)$. Let $P(x_0, y_0)$ be a point on the ellipse. The locus of the intersection of perpendiculars from $A'$ to $AP$ and from $A$ to $A'P$ is given by the equation $\frac{x^2}{a^2} + \frac{y^2}{a^2/b^2} = 1$. Substituting $a^2=2$ and $b^2=1$, we get $\frac{x^2}{2} + \frac{y^2}{2} = 1$, which simplifies to $x^2+y^2=2$. This is conic E, a circle with radius $\sqrt{2}$.

(A) Intersection of ellipse $x^2+2y^2=2$ and conic E $x^2+y^2=2$: Subtracting the equations gives $y^2=0$, so $y=0$. Substituting into $x^2+y^2=2$ gives $x^2=2$, so $x=\pm\sqrt{2}$. The abscissas are $\sqrt{2}$ and $-\sqrt{2}$. Thus, (A) matches with (P) and (Q).

(B) Conic E is $x^2+y^2=2$. This is a circle, which can be considered an ellipse with $a=b=\sqrt{2}$. The length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{2b^2}{a}$. For a circle, $a=b=R=\sqrt{2}$. So, the length of the latus rectum is $\frac{2(\sqrt{2})^2}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$. The square root of this length is $\sqrt{2\sqrt{2}}$. However, if we consider the possibility of a typo and the question meant the length of the latus rectum, it is $2\sqrt{2}$, matching (S).

(C) The foci of the circle $x^2+y^2=2$ are at the center $(0,0)$. The abscissa of the foci is $0$. Thus, (C) matches with (R).

(D) A circle does not have directrices in the standard sense. However, if we consider the ellipse $2x^2+y^2=4$ which is part of the locus derivation, its directrices are $y=\pm 2\sqrt{2}$. If the question implies a directrix of the form $2x+y=k$ related to this ellipse, and if we interpret the coefficients of $x$ and $y$ in the directrix equation as related to the normal vector, then the values $k=\pm 2\sqrt{2}$ might be considered. This is a stretch, but given the options, it's a possible interpretation. Thus, (D) could match with (S) and (T).