Conics Questions (1 Viewer)

lychnobity · Apr 23, 2009

1) Given the ellipse x²/16 + y²/9 = 1. The normal at P(4 cos @, 3 sin @) meets the x-axis in A and the y-axis in B.
Find the locus of the midpoint of AB.

2) Given the hyperbola 2x² - y² = 2 and the circle (x - √3)² + y² = 4.
Find the angle @ if the tangent at P(sec @, √2 tan @) is also a tangent to the circle.

3) 4 points: P, Q, R, S lie on the rectangular hyperbola xy = c². Let their parameters be p, q, r and s respectively. ie P(cp, c/p) etc.
The tangent at P meets the x-axis in A and the y-axis in B.
The tangent at Q is parallel to the tangent at P, and meets the x-axis in C and the y-axis in D.
b) Find the coordinates of A, B, C, D in terms of p.
c) Find the of area of ABCD
d) Find the eqn of the chord PQ
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jet · Apr 23, 2009

I may very well be wrong, but here it is.

$\text{Question 1:} \\ \text{The normal at the point } P(4cos\theta, 3sin\theta) \text{has equation} \\ \frac{4x}{cos\theta} - \frac{3y}{sin\theta} = 7 \\ x = 0, y = -\frac{7sin\theta}{3} \therefore B \equiv (0, -\frac{7sin\theta}{3}) \\ y = 0, x = \frac{7cos\theta}{4} \therefore A \equiv (\frac{7cos\theta}{4}, 0) \\ \therefore \text{The midpoint of AB has } \\ x = \frac{7cos\theta}{8}, cos\theta = \frac{8x}{7} \\ y = -\frac{7sin\theta}{6}, sin\theta = \frac{6y}{7} \\ \text{Hence, } cos^2 \theta + sin^2 \theta = \frac{64x^2}{49} +\frac{36y^2}{49} \\ \frac{64x^2}{49} +\frac{36y^2}{49} = 1 \\ \text{Hence, the locus is an ellipse}$

Dumbledore · Apr 23, 2009

question 2 the algebra gets very messy and i'm incredibly slow at typing, but a way to solve it is find the equation of the tangent to the hyperbola in terms of x,y and @, then solve simultaneously with the circle eliminating y, so u get the equation in terms of x and @ (it will be a quadratic) then use the discriminant and make it = 0 and solve for @.

lychnobity · Apr 23, 2009

jetblack2007 said:
I may very well be wrong, but here it is.

$\text{Question 1:} \\ \text{The normal at the point } P(4cos\theta, 3sin\theta) \text{has equation} \\ \frac{4x}{cos\theta} - \frac{3y}{sin\theta} = 7 \\ x = 0, y = -\frac{7sin\theta}{3} \therefore B \equiv (0, -\frac{7sin\theta}{3}) \\ y = 0, x = \frac{7cos\theta}{4} \therefore A \equiv (\frac{7cos\theta}{4}, 0) \\ \therefore \text{The midpoint of AB has } \\ x = \frac{7cos\theta}{8}, cos\theta = \frac{8x}{7} \\ y = -\frac{7sin\theta}{6}, sin\theta = \frac{6y}{7} \\ \text{Hence, } cos^2 \theta + sin^2 \theta = \frac{64x^2}{49} +\frac{36y^2}{49} \\ \frac{64x^2}{49} +\frac{36y^2}{49} = 1 \\ \text{Hence, the locus is an ellipse}$

Thanks man.

lychnobity · Apr 23, 2009

Dumbledore said:
question 2 the algebra gets very messy and i'm incredibly slow at typing, but a way to solve it is find the equation of the tangent to the hyperbola in terms of x,y and @, then solve simultaneously with the circle eliminating y, so u get the equation in terms of x and @ (it will be a quadratic) then use the discriminant and make it = 0 and solve for @.

Edited the original post, @ is an angle...

Dumbledore · Apr 23, 2009

lychnobity said:
Edited the original post, @ is an angle...

@ is both the parameter and the angle, for example say x^2+y^2=1 is represented by the parametric co ordiantes x=cos@, y= sin@ or P(cos@, sin@)
lets say the angle in the circle is pi/4, you substitue that into the equation as the parameter and you will get x = cos(pi/4) = 1/sqrt(2) and y = sin(pi/4) = 1/sqrt(2) so it also represents the point on the graph. Or you could say it is the angle which subtends a line to the point on the graph.

the method i used does work

jet · Apr 23, 2009

$\text{The tangent at P has equation } x + p^2y = 2cp \\ \therefore A \equiv (2cp ,0), B \equiv (0, \frac{2c}{p}) \\ \text{Now, the tangent at Q has equation } x + q^2y = 2cq \\ \text{but, using the given information, the gradients of the tangents are equal, hence } \\ -\frac{1}{q^2} = -\frac{1}{p^2} \\ \text{Assuming that P and Q are not the same point, we can only deduce that } q = -p \\ \therefore \text{The tangent at Q has equation } x + p^2y = -2cp \\ \therefore C \equiv (-2cp, 0), D \equiv (0, \frac{-2c}{p})$

$\text{It is obvious, since } AB \text{ and } CD \text{ are parallel, and the coordinate axes are perpendicular, that } ABCD \text{ is a rhombus.} \\ \text{Thus, the area } A = \frac{1}{2}AC \cdot BD = \frac{1}{2}(4cp)(\frac{4c}{p}) \\ =8c^2 \\ \text{The chord PQ has gradient } M_{PQ} = \frac{\frac{2c}{p}}{2cp} \\ = \frac{1}{p^2} \\ y - \frac{c}{p} = \frac{1}{p^2}(x - cp) \\ p^2y - cp = x - cp \\ y = \frac{x}{p^2}$

jet · Apr 23, 2009

Dumbledore said:
@ is both the parameter and the angle, for example say x^2+y^2=1 is represented by the parametric co ordiantes x=cos@, y= sin@ or P(cos@, sin@)
lets say the angle in the circle is pi/4, you substitue that into the equation as the parameter and you will get x = cos(pi/4) = 1/sqrt(2) and y = sin(pi/4) = 1/sqrt(2) so it also represents the point on the graph. Or you could say it is the angle which subtends a line to the point on the graph.

the method i used does work

The method you said is the way I was trying it, but the algebra got the best of me and I gave up.

study-freak · Apr 28, 2009

lychnobity said:
2) Given the hyperbola 2x² - y² = 2 and the circle (x - √3)² + y² = 4.

Find the angle @ if the tangent at P(sec @, √2 tan @) is also a tangent to the circle.

Since no one answered this question, I'll post the solution although I guess it's been awhile since this was posted. Btw, LaTeX is so hard to use for me..

$\\$ Tangent at (sec @, 2^{1/2}tan @) $ to 2x^{2}- y^{2}= 2: \\xsec@-\frac{ytan@}{2^{1/2}}=1 \\$If this is a tangent to (x-3^{1/2})^{2}+y^{2}=4, \\$the perpendicular distance from the centre of the circle to the tangent=radius=2 units. Now, Centre(3^{1/2},0)$ and 2^{1/2}xsec@-ytan@-2^{1/2}=0 \\D=\left |\frac{6^{1/2}sec@-2^{1/2}}{(2sec^2@+tan^2@)^{1/2}}\right | =\left |\frac{6^{1/2}sec@-2^{1/2}}{(3sec^2@-1)^{1/2}}\right | =\left |\frac{2^{1/2}(3^{1/2}sec@-1)^{1/2}}{(3^{1/2}sec@+1)^{1/2}} \right |=2 \\$\therefore D^2=\frac{2(3^{1/2}sec@-1)}{3^{1/2}sec@+1}=4 \\$3^{1/2}sec@-1=2(3^{1/2}sec@+1) \\$sec@=-3^{1/2} \\cos@=\frac{-1}{3^{1/2}} \\$\therefore @=\pm (\pi-cos^{-1}(\frac{1}{3^{1/2}}))$ for \pi\geq @>-\pi. \\@\approx \pm 125^{\circ}16'$

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Conics Questions (1 Viewer)

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