Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$

Question

Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ respectively,then find $2M+6m.$

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Let $x=\cos\theta$ and $y=\sin\theta$,because $\sin^2\theta+\cos^2\theta=1$.

Then we need to find the minimum and maximum value of the expression $\frac{4-\sin\theta}{7-\cos\theta}$.

I differentiated it and equated it to zero to find the critical points or points of extrema.

They are $\theta_1=\arcsin(\frac{1}{\sqrt{65}})-\arctan(\frac{7}{4})$ and $\theta_2=\arccos(\frac{1}{\sqrt{65}})+\arctan(\frac{4}{7})$

I found $\frac{4-\sin\theta_1}{7-\cos\theta_1}$ and $\frac{4-\sin\theta_2}{7-\cos\theta_2}$.

$\frac{4-\sin\theta_1}{7-\cos\theta_1}=\frac{3}{4}$ and $\frac{4-\sin\theta_2}{7-\cos\theta_2}=\frac{5}{12}$

This method is full of lengthy calculations.I want to know is there an elegant solution possible for this problem which is short and easy.

Answer 1

$$y-4=z(x-7)\tag{1}$$ is an equation of all the lines that pass $(7,4)$ with slope $z$. When $z$ is at its minimum or maximum then the line touches the unit circle. Equation of tangent lines with slope $z$ to a unit circle is $$y=zx\pm\sqrt{z^2+1}\tag{2}$$ Comparing $(1)$ and $(2)$, $$-7z+4=\pm\sqrt{z^2+1}$$ $$(7z-4)^2=z^2+1$$ $$48z^2-56z+15=0$$ $$(4z-3)(12z-5)=0$$ $$\therefore M=\frac34, m=\frac5{12}$$

Answer 2

HINT:

Let $\dfrac{4-\sin \theta}{7-\cos \theta}=u$

Use Weierstrass Substitution $t=\tan\dfrac\theta2$ to form a Quadratic Equation in $t$ on rearrangement.

As $t$ is real, the discriminant must be non-negative

Answer 3

Let $\displaystyle k=\frac{4-y}{7-x}\Rightarrow 7k-kx=4-y\Rightarrow kx-y = 7k-4$ and given $x^2+y^2=1$

Now using the Cauchy-Schwarz inequality, we get $$[k^2+(-1)^2](x^2+y^2)\geq (kx-y)^2$$

So $$k^2+1\geq (7k-4)^2\Rightarrow 49k^2+16-56k\leq k^2+1$$

So $$48k^2-56k+15\leq 0\Rightarrow (4k-3)(12k-5)\leq 0$$

So we get $\displaystyle \frac{5}{12}\leq k\leq \frac{3}{4}$