JEE Mathematics Ellipse and Hyperbola MCQs Set 07

Answer: (d) \( 9x^2+16y^2=4x^2y^2 \)

Explanation:
1. The tangent at \( \theta \) is \( \frac{x\cos\theta}4+\frac{y\sin\theta}3=1 \), with intercepts \( \frac4{\cos\theta} \) and \( \frac3{\sin\theta} \).
2. The mid-point is \( (h,k)=\left(\frac2{\cos\theta},\frac3{2\sin\theta}\right) \).
3. So \( \frac4{h^2}+\frac9{4k^2}=1 \), i.e. \( 16k^2+9h^2=4h^2k^2 \).

Teacher's Note:
1. Express cos θ and sin θ in terms of h and k and use cos² + sin² = 1.
2. The result is symmetric in a way that 9 goes with x².

Answer: (d) \( 3x-4y=4 \)

Explanation:
1. For mid-point (h, k), the chord is T = S₁: \( 3hx-2ky+2(x+h)-3(y+k)=3h^2-2k^2+4h-6k \).
2. Its slope is \( \frac{3h+2}{2k+3} \). Setting this equal to 2: \( 3h+2=4k+6 \).
3. So \( 3h-4k=4 \) and the locus is \( 3x-4y=4 \).

Teacher's Note:
1. The chord with a given mid-point is T = S₁.
2. Only the coefficients of x and y are needed for the slope.

Answer: (d) infinite

Explanation:
1. Here \( a^2=\cos^2\alpha \) and \( b^2=\sin^2\alpha \). The director circle is \( x^2+y^2=a^2-b^2=\cos2lpha \).
2. For \( \alpha\in(0,\pi/4) \) we have \( \cos2\alpha>0 \), so the director circle is a real circle.
3. Every point of it gives perpendicular tangents, so there are infinitely many such points.

Teacher's Note:
1. The director circle of a hyperbola is x² + y² = a² − b², real only when a > b.
2. a² − b² = cos 2α > 0 here.

Answer: (b) \( y+mx=0 \)

Explanation:
1. For mid-point (h, k), T = S₁ gives \( kx+hy=2hk \). Its slope is \( -\frac kh \).
2. Setting this equal to m gives \( k=-mh \).
3. The locus is \( y=-mx \), i.e. \( y+mx=0 \).

Teacher's Note:
1. For xy = c², the chord with mid-point (h, k) is kx + hy = 2hk.
2. The locus is a diameter through the centre.

Answer: (c) \( \pi/2 \)

Explanation:
1. The hyperbola is \( \frac{x^2}{25}-\frac{y^2}{16}=1 \) with director circle \( x^2+y^2=25-16=9 \).
2. The point satisfies \( 1+8=9 \), so it lies on the director circle.
3. Therefore the tangents from it are perpendicular: \( \pi/2 \).

Teacher's Note:
1. Check whether the point lies on the director circle.
2. The director circle radius² is a² − b².

Answer: (c) a hyperbola with centre \( \left(\frac\alpha2,\frac\beta2\right) \)

Explanation:
1. For mid-point (h, k), T = S₁: \( \frac{hx}{a^2}-\frac{ky}{b^2}=\frac{h^2}{a^2}-\frac{k^2}{b^2} \).
2. The chord passes through \( (\alpha,\beta) \): \( \frac{h\alpha}{a^2}-\frac{k\beta}{b^2}=\frac{h^2}{a^2}-\frac{k^2}{b^2} \).
3. Rearranged: \( \frac{h^2-h\alpha}{a^2}-\frac{k^2-k\beta}{b^2}=0 \), a hyperbola (two lines if degenerate) with centre \( \left(\frac\alpha2,\frac\beta2\right) \).

Teacher's Note:
1. Substitute the fixed point in the chord equation T = S₁.
2. The equation has x² and y² terms of opposite signs.

Answer: (c) \( p^2(a^2\sec^2\alpha+b^2\csc^2\alpha)=(a^2-b^2)^2 \)

Explanation:
1. The normal at \( (a\cos\theta,b\sin\theta) \) is \( \frac{ax}{\cos\theta}-\frac{by}{\sin\theta}=a^2-b^2 \).
2. Compare with \( x\cos\alpha+y\sin\alpha=p \): \( \frac{a}{\cos\theta\cos\alpha}=\frac{-b}{\sin\theta\sin\alpha}=\frac{a^2-b^2}{p} \).
3. So \( \cos\theta=\frac{ap}{(a^2-b^2)\cos\alpha} \) and \( \sin\theta=\frac{-bp}{(a^2-b^2)\sin\alpha} \); squaring and adding gives \( p^2(a^2\sec^2\alpha+b^2\csc^2\alpha)=(a^2-b^2)^2 \).

Teacher's Note:
1. Compare coefficients of the normal with the given line, then use cos² θ + sin² θ = 1.
2. The condition for lx + my = p to be normal is a²/l² + b²/m² = (a² − b²)²/p².

Answer: (d) \( 100x^2y^2=(x^2+y^2)^2(y^2-x^2) \)

Explanation:
1. The normal at \( (5\sec\theta,5\tan\theta) \) is \( x\cos\theta+y\cot\theta=10 \) (using a² + b² = 50 and dividing by 5).
2. The foot of the perpendicular from the origin is \( (h,k)=\frac{10(\cos\theta,\cot\theta)}{\cos^2\theta+\cot^2\theta} \), so \( \frac hk=\frac{\cos\theta}{\cot\theta}=\sin\theta \).
3. Eliminating θ (verified with symbolic computation) gives \( 100x^2y^2=(x^2+y^2)^2(y^2-x^2) \).

Teacher's Note:
1. The normal at the point with parameter θ is ax cos θ + by cot θ = a² + b² for x²/a² − y²/b² = 1.
2. The foot from the origin is the projection onto the line's normal direction.

Answer: (a) 3

Explanation:
1. For any tangent of an ellipse, the product of the perpendicular distances from the two foci equals \( b^2 \).
2. Here \( b^2=3 \).

Teacher's Note:
1. The product of focal perpendiculars on a tangent is b².
2. Check at the vertex: the distances are a − ae and a + ae, whose product is a²(1 − e²) = b².

Answer: (a) 7 units

Explanation:
1. With P = (4 cos θ, 3 sin θ), the normal is \( 4x\sec\theta-3y\csc\theta=7 \) (since a² − b² = 7).
2. Q = (4 cos θ, 4 sin θ), so CQ is the line through the origin along the direction (cos θ, sin θ): points \( (r\cos\theta,\ r\sin\theta) \).
3. Substituting: \( 4r-3r=7 \), so r = 7. CR = a + b = 7.

Teacher's Note:
1. The normal at the eccentric angle θ is ax/cos θ − by/sin θ = a² − b².
2. R lies on the ray from the centre in the direction of Q.

Answer: (d) all of the above

Explanation:
1. A normal of the circle passes through its centre (6, 0). So we need normals of the parabola through (6, 0).
2. The normal with slope m is \( y=-mx+2m+m^3 \); passing through (6, 0): \( 0=-6m+2m+m^3=m(m^2-4) \).
3. So m = 0, 2, −2, giving y = 0, y = −2x + 12 and y = 2x − 12. All three listed lines are common normals.

Teacher's Note:
1. Normals to a circle are the lines through its centre.
2. Find the normals to the parabola from the point (6, 0); there are three.

Answer: (b) \( y=\pm4x\pm\sqrt{37} \)

Explanation:
1. Ellipse: \( \frac{x^2}2+\frac{y^2}5=1 \), tangent condition \( c^2=2m^2+5 \). Hyperbola: \( \frac{x^2}3-\frac{y^2}{11}=1 \), condition \( c^2=3m^2-11 \).
2. Equating: \( 2m^2+5=3m^2-11 \), so \( m^2=16 \) and \( c^2=37 \).
3. The common tangents are \( y=\pm4x\pm\sqrt{37} \).

Teacher's Note:
1. Write both tangent conditions in terms of m and c.
2. Subtract to find m² first.

Answer: (c) \( b=\sqrt3;\ x+2y+4=0 \)

Explanation:
1. The line is \( y=\frac x2+2 \): slope \( \frac12 \), intercept 2 = 1/m, so it touches the parabola.
2. For the ellipse: \( c^2=a^2m^2+b^2 \) gives \( 4=4\cdot\frac14+b^2 \), so \( b^2=3 \).
3. The reflection of the line in the x-axis, \( y=-\frac x2-2 \) i.e. \( x+2y+4=0 \), is also tangent to both the parabola and the ellipse (both curves are symmetric about the x-axis).

Teacher's Note:
1. Use the slope-intercept form and the two tangent conditions.
2. Both curves are symmetric about the x-axis, so the second common tangent is the mirror image.

Answer: (c) a straight line

Explanation:
1. Let the eccentric angles be α and β with α + β = 2γ (constant).
2. The tangents meet at \( \left(\frac{a\cos\frac{\alpha+\beta}2}{\cos\frac{\alpha-\beta}2},\frac{b\sin\frac{\alpha+\beta}2}{\cos\frac{\alpha-\beta}2}\right) \).
3. The ratio \( \frac yx=\frac ba\tan\gamma \) is constant, so the locus is a straight line through the centre.

Teacher's Note:
1. Use the formula for the point of intersection of tangents at two eccentric angles.
2. A constant ratio y/x means a line through the origin.

Answer: (a) \( x^2+y^2=5 \)

Explanation:
1. Here \( a=2,\ b=1 \). The coordinate axes are two perpendicular tangents, so their point of intersection, the origin, lies on the director circle of the ellipse.
2. The director circle has the ellipse's centre as its centre and radius \( \sqrt{a^2+b^2}=\sqrt5 \).
3. So the centre (x, y) is at distance √5 from the origin: \( x^2+y^2=5 \).

Teacher's Note:
1. Perpendicular tangents meet on the director circle.
2. Radius² = a² + b² = 5.

Answer: (c) \( \frac{\sqrt5-1}{2} \)

Explanation:
1. Let Eₙ₋₁ have semi-axes a, b. Then Eₙ has its major axis along the y-axis with focal distance c = b, and it passes through (±a, 0), so its semi-minor axis is a.
2. Its semi-major axis squared is \( a^2+b^2 \), so \( e^2=\frac{b^2}{a^2+b^2} \).
3. With \( b^2=a^2(1-e^2) \): \( e^2=\frac{1-e^2}{2-e^2} \), i.e. \( e^4-3e^2+1=0 \), so \( e^2=\frac{3-\sqrt5}2 \) and \( e=\frac{\sqrt5-1}2 \).

Teacher's Note:
1. Use the focal distance c = ae and the relation a_n² = b_n² + c_n².
2. Solve the quadratic in e².

Answer: (d) \( x^2+y^2=1 \)

Explanation:
1. For conjugate hyperbolas \( \frac1{e^2}+\frac1{e'^2}=1 \) (since \( e^2=1+b^2/a^2 \) and \( e'^2=1+a^2/b^2 \)).
2. So the point (1/e, 1/e′) satisfies \( x^2+y^2=1 \).

Teacher's Note:
1. Compute 1/e² = a²/(a² + b²) and 1/e′² = b²/(a² + b²).
2. They add to 1.

Answer: (d) \( a^2 \)

Explanation:
1. Let P = (x₁, y₁). Then ON = x₁.
2. The tangent \( \frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1 \) meets y = 0 at \( x=\frac{a^2}{x_1} \), so OT = a²/x₁.
3. Therefore OT · ON = a².

Teacher's Note:
1. Put y = 0 in the tangent at (x₁, y₁).
2. ON is just the x-coordinate of P.

Answer: (a) \( \frac{a^2}{x^2}-\frac{b^2}{y^2}=1 \)

Explanation:
1. Let P = (x₁, y₁). The tangent cuts the axes at \( A\left(\frac{a^2}{x_1},0\right) \) and \( B\left(0,-\frac{b^2}{y_1}\right) \).
2. The vertex of the rectangle opposite O is \( Q=\left(\frac{a^2}{x_1},-\frac{b^2}{y_1}\right) \), so \( x_1=\frac{a^2}{X},\ y_1=-\frac{b^2}{Y} \).
3. Since P is on the hyperbola: \( \frac{a^2}{X^2}-\frac{b^2}{Y^2}=1 \).

Teacher's Note:
1. Find the intercepts of the tangent, then the opposite vertex of the rectangle with O.
2. Substitute x₁ and y₁ in the hyperbola equation.

Answer: (b) \( -\frac{a^4}{b^4} \)

Explanation:
1. The chords of contact are \( \frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1 \) and \( \frac{xx_2}{a^2}-\frac{yy_2}{b^2}=1 \) with slopes \( \frac{b^2x_1}{a^2y_1} \) and \( \frac{b^2x_2}{a^2y_2} \).
2. Perpendicular: \( \frac{b^4x_1x_2}{a^4y_1y_2}=-1 \).
3. So \( \frac{x_1x_2}{y_1y_2}=-\frac{a^4}{b^4} \).

Teacher's Note:
1. The chord of contact from (x₁, y₁) is T = 0.
2. Product of slopes = −1.

Answer: (c) \( \frac{x^2}9-\frac{y^2}4=\left(\frac{x^2+y^2}9\right)^2 \)

Explanation:
1. Let the point on the hyperbola be (x₁, y₁); the chord of contact to the circle is \( xx_1+yy_1=9 \).
2. The mid-point (h, k) of a chord of the circle satisfies \( hx+ky=h^2+k^2 \). Comparing: \( x_1=\frac{9h}{h^2+k^2},\ y_1=\frac{9k}{h^2+k^2} \).
3. Substituting in the hyperbola: \( \frac{9h^2}{(h^2+k^2)^2}-\frac{81k^2}{4(h^2+k^2)^2}=1 \), which simplifies to \( \frac{h^2}9-\frac{k^2}4=\left(\frac{h^2+k^2}{9}\right)^2 \).

Teacher's Note:
1. The chord of the circle with mid-point (h, k) is T = S₁.
2. Match coefficients to express x₁ and y₁ through h and k.

Answer: (c) \( \frac{(x-5)^2}{25}+\frac{y^2}{16}=1 \)

Explanation:
1. Use \( \tan\frac B2\tan\frac C2=\frac{s-a}{s} \) with a = BC = 6. The condition gives \( \frac{s-a}s=\frac14 \), so \( s=\frac{4a}3=8 \).
2. Then \( b+c=2s-a=10 \): the sum of distances AB + AC is constant (10) > BC = 6.
3. So A lies on an ellipse with foci B and C, \( 2a'=10,\ c'=3,\ b'=4 \) and centre (5, 0): \( \frac{(x-5)^2}{25}+\frac{y^2}{16}=1 \).

Teacher's Note:
1. tan(B/2)tan(C/2) = (s − a)/s in any triangle.
2. A constant sum of distances to two fixed points defines an ellipse.