Conic Sections

Let P : y² = 4ax, a > 0 be a parabola with focus S. Let the tangents to the parabola P make an angle of $\frac{π}{4}$ with the line y = 3x + 5 touch the parabola P at A and B. Then the value of a for which A, B and S are collinear is

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alok kumar singh

Contributor-Level 10

Tangents making angle $\frac{π}{4}$ with y = 3x + 5.

$t a n \frac{π}{4} = | \frac{m - 3}{1 + 3 m} | \Rightarrow m = - 2, \frac{1}{2}$

So, these tangents are . So ASB is a focal chord.

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2 weeks ago

If the tangents drawn at the points O(0, 0) and $P (1 + \sqrt[]{5}, 2)$ on the circle x² + y² – 2x – 4y = 0 intersect at the point Q, then the area of the triangle OPQ is equal to

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Contributor-Level 9

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2 weeks ago

Let the eccentricity of the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 b e \sqrt[]{\frac{5}{2}}$ and length of its latus rectum be $6 \sqrt[]{2}, i f y = 2 x + c$ is a tangent to the hyperbola H, then the value of c² is equal to

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Contributor-Level 9

$a^{2} (e^{2} - 1) = b^{2}$

$e = \sqrt[]{\frac{5}{2}} \Rightarrow b^{2} = \frac{3 a^{2}}{2}$

Length of latus rectum $\frac{2 b^{2}}{a} = 6 \sqrt[]{2}$

$3 a = 6 \sqrt[]{2} \Rightarrow a = 2 \sqrt[]{2}$

$b = 2 \sqrt[]{3}$

y = 2x + c is tangent to hyperbola

$∴ c^{2} = a^{2} m^{2} - b^{2} = 20$

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2 weeks ago

Let the tangent drawn to the parabola y² = 24x at the point (a, b) is perpendicular to the line 2x + 2y = 5. Then the normal to the hyperbola $\frac{x^{2}}{α^{2}} - \frac{y^{2}}{β^{2}} = 1$ at the point (a + 4, b + 4) does NOT pass through the point:

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Contributor-Level 9

Any tangent to y² = 24x at (a, b) is by = 12 (x + a) therefore Slope = $\frac{12}{β}$

and perpendicular to 2x + 2y = 5 Þ 12 = b and a = 6 Hence hyperbola is $\frac{x^{2}}{6^{2}} - \frac{y^{2}}{1 2^{2}}$ = 1 and normal is drawn at (10, 16)

therefore equation of normal $\frac{36 x}{10} + \frac{144 y}{16} = 36 + 144 \Rightarrow \frac{x}{50} + \frac{y}{20} = 1$ This does not pass through (15, 13) out of given option.

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3 weeks ago

If the co-ordinate of the vertex of the vertex of the parabola whose parametric equation is x = t² – t + 1 and y = t² + t + 1, t $\in$ R is (a, b) then (2a + 4b) equals

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Contributor-Level 10

x = t² – t + 1 … (1)

y = t² + t + 1 … (2)

y – x = 2t & x + y = 2 (t² + 1)

__________on eliminating 't' we get

$\Rightarrow (x + y - 2) = 2 {(\frac{y - x}{2})}^{2}$

${(x - y)}^{2} = 2 (x + y - 2)$

Axis : x – y = 0

Tangent at vertex : x + y – 2 = 0

Vertex : (1, 1) = (x, y)

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3 weeks ago

Conic represented by x = 4(cost – sint) and y = 5(cost + sint) is (where t is the parameter)

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Contributor-Level 10

$\frac{x}{4} = c o s t - s i n t . . . (1)$

$\frac{y}{5} = c o s t + s i n t$ . . . . (2)

(1)² + (2)² gives

${(\frac{x}{4})}^{2} + {(\frac{y}{5})}^{2} = 2$

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3 weeks ago

Let x² + y² – 4x – 2y – 11 = 0 be a circle. A pair of tangents from point P(4, 5) with a pair of raii form a quadrilateral of area

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Contributor-Level 10

area of quad PACB = LR = 8

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3 weeks ago

Find minimum distance between the circle x² + (y-3)² = 1 and parabola y² = 4x

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Contributor-Level 9

Equation of normal $y = m x - 2 m - m^{3}$ must pass through centre $(0,3)$

$\begin{array}{r} m^{3} + 2 m + 3 = 0 \\ m^{2} (m + 1) - m (m + 1) + 3 \cdot 1 (m + 1) = 0 \\ (m + 1) (m^{2} - m + 3) = 0 \\ m = - 1 \end{array}$

$∴$ point of contact of normal at parabola is $({a m}^{2}, - 2 a m) = (1,2)$

So distance between parabola and circle is $= \sqrt[]{2} - 1$

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New answer posted

3 weeks ago

A chord is drawn from P(0, 4) to the ellipse x²/a² + y²/b² = 1, (a>b, b<2) which intersects the major axis at M, foot of perpendicular from centre of ellipse to PM is N, Find PM.NP

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Contributor-Level 9

Clearly PM. PN $= {O P}^{2}$ =OP2

i.e. $P M \cdot P N = 16$ PM⋅PN=16

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3 weeks ago

If x&y are real numbers satisfying the relation x² + y² – 6x + 8y + 24 = 0 then minimum value of log₂(x² + y²) is

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