The equation of a parabola is given by the definition: the distance from any point on the parabola to the focus is equal to the perpendicular distance from the point to the directrix.
Given:
- Focus \( F = (-2, 1) \)
- Directrix: \( 2x + y + 2 = 0 \)
The distance from the point \( P(x_1, y_1) \) on the parabola to the focus is: \[ \text{Distance to focus} = \sqrt{(x_1 + 2)^2 + (y_1 - 1)^2} \] The distance from \( P(x_1, y_1) \) to the directrix \( 2x + y + 2 = 0 \) is: \[ \text{Distance to directrix} = \frac{|2x_1 + y_1 + 2|}{\sqrt{2^2 + 1^2}} = \frac{|2x_1 + y_1 + 2|}{\sqrt{5}} \] For \( x_1 = -2 \), we substitute \( x_1 = -2 \) into both expressions: \[ \sqrt{(-2 + 2)^2 + (y_1 - 1)^2} = \frac{|2(-2) + y_1 + 2|}{\sqrt{5}} \] Simplifying both sides, we solve for \( y_1 \). After solving, we find: \[ y_1 = \frac{3}{2} \]
Thus, the sum of the ordinates of the points on the parabola is \( \frac{3}{2} \).
Equation of the parabola is given by: \[ (x + 2)^2 + (y - 1)^2 = \left(\frac{2x + y + 2}{\sqrt{5}}\right)^2 \] Multiplying both sides by 5: \[ 5[(x + 2)^2 + (y - 1)^2] = (2x + y + 2)^2 \] Substitute \(x = -2\): \[ 5(y - 1)^2 = (y - 2)^2 \] Expanding both sides: \[ 5(y^2 - 2y + 1) = y^2 - 4y + 4 \] \[ 5y^2 - 10y + 5 = y^2 - 4y + 4 \] \[ 4y^2 - 6y + 1 = 0 \] Sum of roots: \[ y_1 + y_2 = \frac{6}{4} = \frac{3}{2} \] \[ \boxed{y_1 + y_2 = \frac{3}{2}} \]
Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
Given below are two statements:
Statement (I):
are isomeric compounds.
Statement (II):
are functional group isomers.
In the light of the above statements, choose the correct answer from the options given below:
Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with
\(K_4\)[Fe(CN)\(_6\)] is : \[ {Cu}^{2+}, \, {Fe}^{3+}, \, {Ba}^{2+}, \, {Ca}^{2+}, \, {NH}_4^+, \, {Mg}^{2+}, \, {Zn}^{2+} \]