Let the product of the focal distances of the point P(4, 2√(3))$P(4, 2\sqrt{3})$ on the hyperbola H : (x²)/(a²) - (y²)/(b²) = 1$H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be 32[cite: 690, 691]. Let the length of the conjugate axis of H$H$ be p$p$ and the length of its latus rectum be q$q$[cite: 692]. Then p² + q²$p^2 + q^2$ is equal to[cite: 693]:

Related Formula

The equation of a family of curves passing through the intersection of two conics S₁ = 0$S_1 = 0$ and S₂ = 0$S_2 = 0$ is S₁ + λ S₂ = 0$S_1 + \lambda S_2 = 0$. For this resulting curve to be a circle, the coefficient of x²$x^2$ must equal the coefficient of y²$y^2$, and the coefficient of the xy$xy$ term must be zero.

Core Logic

Let the two ellipses be: S₁ ≡ x² + 2y² - 6x - 12y + 23 = 0$S_1 \equiv x^2 + 2y^2 - 6x - 12y + 23 = 0$ S₂ ≡ 4x² + 2y² - 20x - 12y + 35 = 0$S_2 \equiv 4x^2 + 2y^2 - 20x - 12y + 35 = 0$

Equation of the curve passing through their intersection is S₁ + λ S₂ = 0$S_1 + \lambda S_2 = 0$:

For this to represent a circle, coefficient of x²$x^2$ = coefficient of y²$y^2$:

Step 1: Finding Circle Parameters

Substitute λ = 1/2$\lambda = 1/2$ back into the family equation:

Dividing by 3 to write in standard form:

The centre (a, b)$(a, b)$ is given by (-g, -f)$(-g, -f)$:

The radius r$r$ is given by r² = g² + f² - c$r^2 = g^2 + f^2 - c$:

Step 2: Final Calculation

We need to find ab + 18r²$ab + 18r^2$:

Pattern Recognition

When intersection points of two 2nd$2^{\text{nd}}$ degree curves form a circle, apply S₁ + λ S₂ = 0$S_1 + \lambda S_2 = 0$ immediately, forcing the necessary symmetric coefficients to extract λ$\lambda$.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Circles

Related Formula

Eccentricity of a hyperbola (x²)/(a²) - (y²)/(b²) = 1$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is e = √(1 + (b²)/(a²))$e = \sqrt{1 + \frac{b^2}{a^2}}$ Parametric coordinates on a hyperbola are (a θ, b θ)$(a\sec\theta, b\tan\theta)$.

Core Logic

The equation is (x²)/(4) - (y²)/(8) = 1$\frac{x^2}{4} - \frac{y^2}{8} = 1$.

Let the point P$P$ on the hyperbola be (2 θ, 2√(2) θ)$(2\sec\theta, 2\sqrt{2}\tan\theta)$. Since chord PQ$PQ$ is perpendicular to the x-axis, the triangle OPQ$OPQ$ has the x-axis as its altitude OM$OM$. In equilateral OPQ$\triangle OPQ$, half the vertex angle is 30^°$30^\circ$:

Step 1: Calculating Coordinates

Step 2: Area Calculation

The area of OPQ = 2 × ((1)/(2) × OM × PM) = OM × PM$\triangle OPQ = 2 \times \left(\frac{1}{2} \times OM \times PM\right) = OM \times PM$

Since θ = 1√(6)$\sin\theta = \frac{1}{\sqrt{6}}$, we have ²θ = 1 - (1)/(6) = (5)/(6)$\cos^2\theta = 1 - \frac{1}{6} = \frac{5}{6}$.

Pattern Recognition

A perpendicular chord symmetric about the principal axis automatically splits into two right triangles. Utilizing the parametric coordinates directly defines the ratio of sides matching 30^°$\tan 30^\circ$.

Chapter Mix

Class 11 Maths: Hyperbola

Related Formula

Parametric form of y² = 4ax$y^2 = 4ax$ is (at², 2at)$(at^2, 2at)$. Minimum distance from origin to a circle with centre C$C$ and radius r$r$ is |OC - r|$|OC - r|$.

Core Logic

Step 1: Defining the Circle

Vertices are A((2√(3))², 2(2√(3))) = A(12, 4√(3))$A((2\sqrt{3})^2, 2(2\sqrt{3})) = A(12, 4\sqrt{3})$ and B(12, -4√(3))$B(12, -4\sqrt{3})$. The circle has diameter AB$AB$. Therefore, its centre is the midpoint of AB$AB$, which is C(12, 0)$C(12, 0)$. The radius R$R$ is half the length of AB$AB$, so R = 4√(3)$R = 4\sqrt{3}$.

The equation of the circle is:

Step 2: Calculating Distance

The distance from the origin O(0,0)$O(0,0)$ to the centre C(12, 0)$C(12, 0)$ is d = 12$d = 12$. The minimum distance from the origin to the circle is |d - R|$|d - R|$:

Pattern Recognition

Any polygon inscribed in a standard parabola symmetric across the axis can be defined completely by equating the slope of a vertex from the origin to the respective tangent of the split angle.

Chapter Mix

Class 11 Maths: Parabola Class 11 Maths: Circles

Related Formula

Core Logic

Step 1: Distance and Coordinate Mapping

AB² = 64$AB^2 = 64$

Step 2: Locus of Centroid

Let the centroid be G(h,k)$G(h,k)$.

Substituting a$a$ and b$b$ into the equation:

Step 3: Radius of Locus

This is a circle with r² = (64)/(9)$r^2 = \frac{64}{9}$, so r = (8)/(3)$r = \frac{8}{3}$.

Pattern Recognition

Whenever a circle passes through the origin and points on the axes, the segment joining the axes points is always the diameter. Centroid coordinates easily scale the locus.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines

Related Formula

Core Logic

For E₁$E_1$: a>b ⇒ e = (4)/(5)$a>b \Rightarrow e = \frac{4}{5}$. Distance between foci 2ae = 8 ⇒ a(4/5) = 4 ⇒ a = 5$2ae = 8 \Rightarrow a(4/5) = 4 \Rightarrow a = 5$.

₁ = (2b²)/(a) = (18)/(5)$\ell_1 = \frac{2b^2}{a} = \frac{18}{5}$

Step 1: Analysing E2

For E₂$E_2$: A$A Vertical ellipse. e = (4)/(5)$e = \frac{4}{5}$.