The focus of the parabola y²=4x+16 is the centre of the circle C of radius 5. If the values of λ, fo

Q: Why do we need to rewrite the parabola equation? Can't we just find the focus from $y^{2} = 4x + 16$ directly?

While you could technically derive it, rewriting it into the standard form $Y^{2} = 4aX$ is the safest way to avoid sign errors. By shifting the origin to the vertex $(-4, 0)$, you clearly see that $a=1$, making the focus calculation $F = (-4 + 1, 0) = (-3, 0)$ much more intuitive and less prone to mistakes.

Q: Is there a faster way to solve for the intersection point $P$ without using substitution?

You could use Cramer's rule or matrix inversion, but for a system of two linear equations, substitution is usually the most direct path. Since $y = 3x$ is given, substituting it into the second equation $x + \lambda y = 4$ immediately isolates $x$, which is the most efficient algebraic route.

Q: How do I know which value is $\lambda_{1}$ and which is $\lambda_{2}$?

The problem explicitly states the condition $\lambda_{1} < \lambda_{2}$. Once you solve the quadratic equation $6\lambda^{2} + \lambda - 7 = 0$ and find the two roots, simply compare their numerical values. The smaller root is $\lambda_{1}$ and the larger one is $\lambda_{2}$.

Question

The focus of the parabola $y^{2}=4x+16$ is the centre of the circle C of radius 5. If the values of $\lambda$, for which C passes through the point of intersection of the lines $3x-y=0$ and $x+\lambda y=4$ are $\lambda_{1}$ and $\lambda_{2}$, $\lambda_{1}<\lambda_{2}$, then $12\lambda_{1}+29\lambda_{2}$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The parabola is defined by the equation $y^{2} = 4x + 16$. To understand its geometry, we factor out the coefficient of $x$ to reach the standard form:

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

This represents a parabola with its vertex at $(-4, 0)$ and a parameter $a = 1$. The focus $F$ of the parabola lies $a$ units to the right of the vertex, placing it at $(-3, 0)$.

Given that the focus $F$ is the center of circle $C$ with a radius of $5$, the equation of the circle is:

$$(x + 3)^{2} + y^{2} = 25$$

### The Intersection Dance

We consider two lines, $3x - y = 0$ and $x + \lambda y = 4$, which intersect at point $P$. From the first line, we have the relationship $y = 3x$.

Substituting this into the second line, we obtain $x + \lambda(3x) = 4$, which simplifies to $x(1 + 3\lambda) = 4$. Thus, the coordinates of the intersection point $P$ are:

$$x = \frac{4}{1 + 3\lambda}, \quad y = \frac{12}{1 + 3\lambda}$$

### The Final Convergence

Since the circle $C$ passes through point $P$, the coordinates of $P$ must satisfy the circle's equation:

$$\left(\frac{4}{1 + 3\lambda} + 3\right)^{2} + \left(\frac{12}{1 + 3\lambda}\right)^{2} = 25$$

Simplifying the first term inside the square, we get $\frac{4 + 3(1 + 3\lambda)}{1 + 3\lambda} = \frac{7 + 9\lambda}{1 + 3\lambda}$. The equation becomes:

$$\frac{(7 + 9\lambda)^{2}}{(1 + 3\lambda)^{2}} + \frac{144}{(1 + 3\lambda)^{2}} = 25$$

Multiplying by $(1 + 3\lambda)^{2}$ to clear the fractions, we have:

$$(7 + 9\lambda)^{2} + 144 = 25(1 + 3\lambda)^{2}$$

Expanding both sides yields:

$$49 + 126\lambda + 81\lambda^{2} + 144 = 25(1 + 6\lambda + 9\lambda^{2})$$

$$81\lambda^{2} + 126\lambda + 193 = 225\lambda^{2} + 150\lambda + 25$$

Rearranging all terms to one side results in the quadratic equation:

$$144\lambda^{2} + 24\lambda - 168 = 0$$

Dividing by $24$, we obtain:

$$6\lambda^{2} + \lambda - 7 = 0$$

Factoring the quadratic gives $(6\lambda + 7)(\lambda - 1) = 0$, yielding roots $\lambda = 1$ and $\lambda = -\frac{7}{6}$. Given $\lambda_{1} < \lambda_{2}$, we identify $\lambda_{1} = -\frac{7}{6}$ and $\lambda_{2} = 1$.

Calculating the final value $12\lambda_{1} + 29\lambda_{2}$:

$$12\left(-\frac{7}{6}\right) + 29(1) = -14 + 29 = 15$$

The final result is **15**.

Enter Numerical Value:

Visualized Solution

Analyze the Parabola $y^{2} = 4 x + 16$

Locate the Focus $F$

Define the Circle $C$

Intersection of the Two Lines

Solve for Intersection Point $P$

Substitute $P$ into Circle

Simplify the Expression

Expand and Rearrange

Form the Quadratic Equation

Solve for $λ$

Final Calculation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Intersection Dance

The Final Convergence

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